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All questions of Chapter 15: Probability for CA Foundation Exam
What is the chance of getting at least one defective item if 3 items are drawn randomly from a lot containing 6 items of which 2 are defective item?- a)0.30
- b)0.20
- c)0.80
- d)0.50
Correct answer is option 'C'. Can you explain this answer?
What is the chance of getting at least one defective item if 3 items are drawn randomly from a lot containing 6 items of which 2 are defective item?
a)
0.30
b)
0.20
c)
0.80
d)
0.50
|
Lakshmi Kaur answered |
Solution:
To find the probability of getting at least one defective item, we need to find the probability of getting 1, 2, or 3 defective items.
Probability of getting 1 defective item out of 3 drawn:
The first defective item can be drawn in 2C1 ways (selecting 1 item out of 2 defective items), and the remaining 2 non-defective items can be drawn in 4C2 ways (selecting 2 items out of 4 non-defective items). Therefore, the probability of getting 1 defective item is:
P(1) = (2C1 x 4C2) / 6C3 = (2 x 6) / 20 = 0.6
Probability of getting 2 defective items out of 3 drawn:
The two defective items can be drawn in 2C2 ways (selecting 2 items out of 2 defective items), and the remaining 1 non-defective item can be drawn in 4C1 ways (selecting 1 item out of 4 non-defective items). Therefore, the probability of getting 2 defective items is:
P(2) = (2C2 x 4C1) / 6C3 = (1 x 4) / 20 = 0.2
Probability of getting 3 defective items out of 3 drawn:
The three defective items can be drawn in 2C3 ways (selecting 3 items out of 2 defective items, which is not possible), so the probability of getting 3 defective items is 0.
Therefore, the probability of getting at least one defective item is:
P(at least 1) = P(1) + P(2) + P(3) = 0.6 + 0.2 + 0 = 0.8
Hence, the correct answer is option C, 0.8.
To find the probability of getting at least one defective item, we need to find the probability of getting 1, 2, or 3 defective items.
Probability of getting 1 defective item out of 3 drawn:
The first defective item can be drawn in 2C1 ways (selecting 1 item out of 2 defective items), and the remaining 2 non-defective items can be drawn in 4C2 ways (selecting 2 items out of 4 non-defective items). Therefore, the probability of getting 1 defective item is:
P(1) = (2C1 x 4C2) / 6C3 = (2 x 6) / 20 = 0.6
Probability of getting 2 defective items out of 3 drawn:
The two defective items can be drawn in 2C2 ways (selecting 2 items out of 2 defective items), and the remaining 1 non-defective item can be drawn in 4C1 ways (selecting 1 item out of 4 non-defective items). Therefore, the probability of getting 2 defective items is:
P(2) = (2C2 x 4C1) / 6C3 = (1 x 4) / 20 = 0.2
Probability of getting 3 defective items out of 3 drawn:
The three defective items can be drawn in 2C3 ways (selecting 3 items out of 2 defective items, which is not possible), so the probability of getting 3 defective items is 0.
Therefore, the probability of getting at least one defective item is:
P(at least 1) = P(1) + P(2) + P(3) = 0.6 + 0.2 + 0 = 0.8
Hence, the correct answer is option C, 0.8.
When 3 unbiased coins are tossed. The probability of obtaining 3 heads is- a)2/4
- b)¼
- c)¾
- d)0
Correct answer is option 'D'. Can you explain this answer?
When 3 unbiased coins are tossed. The probability of obtaining 3 heads is
a)
2/4
b)
¼
c)
¾
d)
0
|
Poonam Reddy answered |
Correct Answer :- d
Explanation :
Total no. Of outcomes: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
No. Outcomes of all heads: 1
Prob. Of all heads = no. Of outcomes of all heads/ total no. Of outcomes
i.e (1/8)
Four married couples have gathered in a room. Two persons are selected at random amongst them, find the probability that selected persons are a gentleman and a lady but not a Couple.- a)1/7
- b)3/7
- c)1/8
- d)3/8
Correct answer is option 'B'. Can you explain this answer?
Four married couples have gathered in a room. Two persons are selected at random amongst them, find the probability that selected persons are a gentleman and a lady but not a Couple.
a)
1/7
b)
3/7
c)
1/8
d)
3/8
|
|
Jyoti Nair answered |
Solution:
Given that there are 4 married couples in the room.
We need to find the probability that the selected persons are a gentleman and a lady but not a couple.
Let's consider the total number of ways in which we can select 2 persons from 8 persons.
Total number of ways = 8C2 = (8 x 7) / (2 x 1) = 28
Case 1: Selecting a gentleman and a lady who are not a couple.
There are 4 gentlemen and 4 ladies in the room.
We can select 1 gentleman out of 4 in 4C1 ways.
We can select 1 lady out of 4 in 4C1 ways.
Out of these 2 selected persons, they should not be a couple.
There are 3 ladies for each gentleman who is not his wife.
Therefore, each gentleman has 3 choices for a lady who is not his wife.
So, the number of ways to select a gentleman and a lady who are not a couple = 4C1 x 3C1 = 12
Therefore, the probability of selecting a gentleman and a lady who are not a couple = (12/28)
Case 2: Selecting a gentleman and a lady who are a couple.
There are 4 couples in the room.
We can select 1 couple out of 4 in 4C1 ways.
We cannot select any other person along with them.
Therefore, the number of ways to select a gentleman and a lady who are a couple = 4C1 = 4
Therefore, the probability of selecting a gentleman and a lady who are a couple = (4/28)
Total probability of selecting a gentleman and a lady = Probability of case 1 + Probability of case 2
= (12/28) + (4/28)
= (16/28)
= (4/7)
Therefore, the probability of selecting a gentleman and a lady who are not a couple = (4/7) or option B.
Given that there are 4 married couples in the room.
We need to find the probability that the selected persons are a gentleman and a lady but not a couple.
Let's consider the total number of ways in which we can select 2 persons from 8 persons.
Total number of ways = 8C2 = (8 x 7) / (2 x 1) = 28
Case 1: Selecting a gentleman and a lady who are not a couple.
There are 4 gentlemen and 4 ladies in the room.
We can select 1 gentleman out of 4 in 4C1 ways.
We can select 1 lady out of 4 in 4C1 ways.
Out of these 2 selected persons, they should not be a couple.
There are 3 ladies for each gentleman who is not his wife.
Therefore, each gentleman has 3 choices for a lady who is not his wife.
So, the number of ways to select a gentleman and a lady who are not a couple = 4C1 x 3C1 = 12
Therefore, the probability of selecting a gentleman and a lady who are not a couple = (12/28)
Case 2: Selecting a gentleman and a lady who are a couple.
There are 4 couples in the room.
We can select 1 couple out of 4 in 4C1 ways.
We cannot select any other person along with them.
Therefore, the number of ways to select a gentleman and a lady who are a couple = 4C1 = 4
Therefore, the probability of selecting a gentleman and a lady who are a couple = (4/28)
Total probability of selecting a gentleman and a lady = Probability of case 1 + Probability of case 2
= (12/28) + (4/28)
= (16/28)
= (4/7)
Therefore, the probability of selecting a gentleman and a lady who are not a couple = (4/7) or option B.
The probability of winning of a person is 6/11 and at a result he gets Rs.77/= .The expectation of this person is
- a)Rs.35/=
- b)Rs.42/=
- c)Rs.58/=
- d)none
Correct answer is option 'B'. Can you explain this answer?
The probability of winning of a person is 6/11 and at a result he gets Rs.77/= .The expectation of this person is
a)
Rs.35/=
b)
Rs.42/=
c)
Rs.58/=
d)
none
|
Rajveer Jain answered |
Solution:
Given, probability of winning of a person = 6/11
Amount received by the person = Rs. 77/=
To find: Expectation of the person
Calculation:
The formula to find the expectation of a person is:
Expectation = (Amount received x Probability of winning) - (Amount lost x Probability of losing)
Here, the probability of losing = 1 - Probability of winning
= 1 - (6/11)
= 5/11
Substituting the given values in the formula, we get:
Expectation = (77 x 6/11) - (0 x 5/11)
= 42
Therefore, the expectation of the person is Rs.42/=. Hence, option B is the correct answer.
Conclusion:
The expectation of the person who wins with a probability of 6/11 and receives Rs.77/= is Rs.42/=.
Given, probability of winning of a person = 6/11
Amount received by the person = Rs. 77/=
To find: Expectation of the person
Calculation:
The formula to find the expectation of a person is:
Expectation = (Amount received x Probability of winning) - (Amount lost x Probability of losing)
Here, the probability of losing = 1 - Probability of winning
= 1 - (6/11)
= 5/11
Substituting the given values in the formula, we get:
Expectation = (77 x 6/11) - (0 x 5/11)
= 42
Therefore, the expectation of the person is Rs.42/=. Hence, option B is the correct answer.
Conclusion:
The expectation of the person who wins with a probability of 6/11 and receives Rs.77/= is Rs.42/=.
Out of the following values, which one is not possible in probability?- a)P(x) = 1
- b)∑ x P(x) = 3
- c)P(x) = 0.5
- d)P(x) = – 0.5
Correct answer is option 'D'. Can you explain this answer?
Out of the following values, which one is not possible in probability?
a)
P(x) = 1
b)
∑ x P(x) = 3
c)
P(x) = 0.5
d)
P(x) = – 0.5
|
Freedom Institute answered |
In probability P(x) is always greater than or equal to zero.
Probability of getting a head when two unbiased coins are tossed simultaneously is
- a)0.25
- b)0.5
- c)0.20
- d)0.75
Correct answer is option 'B'. Can you explain this answer?
Probability of getting a head when two unbiased coins are tossed simultaneously is
a)
0.25
b)
0.5
c)
0.20
d)
0.75
|
Srsps answered |
Favourable number of outcomes, getting one head ={HT, TH}
Total number of outcomes ={HH, HT, TH, TT}
Probability =2/4=1/2
Total number of outcomes ={HH, HT, TH, TT}
Probability =2/4=1/2
The probability of a cricket term winning match at Kanpur is 2/5 and hosing match at Delhi is 1/7 what is the Probability of the term winning atleast one match?- a)3/35
- b)32/35
- c)18/35
- d)17/35
Correct answer is option 'B'. Can you explain this answer?
The probability of a cricket term winning match at Kanpur is 2/5 and hosing match at Delhi is 1/7 what is the Probability of the term winning atleast one match?
a)
3/35
b)
32/35
c)
18/35
d)
17/35
|
|
Freedom Institute answered |
B is correct.
Step-by-step explanation:
Probability( winning a match at Kanpur),P(WK) 
Probability( losing a match at Kanpur),P(LK)
Probability( losing a match at Delhi), P(LD)
Probability( winning a match at Delhi),P(WD)
The probability of the team winning at least one match,
P( team winning at least one match ) =1 - P(Losing both math)
Probability( losing a match at Kanpur),P(LK)
Probability( losing a match at Delhi), P(LD)
Probability( winning a match at Delhi),P(WD)
The probability of the team winning at least one match,
P( team winning at least one match ) =1 - P(Losing both math)
Hence, the probability of the team winning at least one match is 
A bag contains 12 balls which are numbered from 1 to 12. If a ball is selected at random, what is the probability that the number of the ball will be a multiple of 5 or 6 ?- a)0.30
- b)0.25
- c)0.20
- d)1/3
Correct answer is option 'D'. Can you explain this answer?
A bag contains 12 balls which are numbered from 1 to 12. If a ball is selected at random, what is the probability that the number of the ball will be a multiple of 5 or 6 ?
a)
0.30
b)
0.25
c)
0.20
d)
1/3
|
|
Meera Basak answered |
Solution:
Multiples of 5: 5, 10
Multiples of 6: 6, 12
The numbers which are multiple of 5 or 6 are: 5, 6, 10, 12
Total number of balls = 12
Therefore, the required probability is:
P(multiple of 5 or 6) = (number of favorable outcomes)/(total number of outcomes)
P(multiple of 5 or 6) = 4/12
P(multiple of 5 or 6) = 1/3
Therefore, the correct option is (d) 1/3.
Multiples of 5: 5, 10
Multiples of 6: 6, 12
The numbers which are multiple of 5 or 6 are: 5, 6, 10, 12
Total number of balls = 12
Therefore, the required probability is:
P(multiple of 5 or 6) = (number of favorable outcomes)/(total number of outcomes)
P(multiple of 5 or 6) = 4/12
P(multiple of 5 or 6) = 1/3
Therefore, the correct option is (d) 1/3.
In a class 40 % students read Mathematics, 25 % Biology and 15 % both Mathematics and Biology. One student is select at random. The probability that he reads Mathematics if it is known that he reads Biology is- a)2/5
- b)3/5
- c)4/5
- d)none
Correct answer is option 'B'. Can you explain this answer?
In a class 40 % students read Mathematics, 25 % Biology and 15 % both Mathematics and Biology. One student is select at random. The probability that he reads Mathematics if it is known that he reads Biology is
a)
2/5
b)
3/5
c)
4/5
d)
none
|
|
Sonal Patel answered |
Given information:
- 40% of students read Mathematics.
- 25% of students read Biology.
- 15% of students read both Mathematics and Biology.
To find:
- The probability that a student reads Mathematics if it is known that he reads Biology.
Solution:
Step 1: Draw a Venn diagram representing the given information.
The rectangle represents the total number of students in the class. The circle on the left represents the number of students who read Mathematics, and the circle on the right represents the number of students who read Biology. The overlapping region represents the number of students who read both Mathematics and Biology.
Step 2: Calculate the number of students who read only Biology.
To do this, we subtract the number of students who read both Mathematics and Biology from the total number of students who read Biology.
Number of students who read only Biology = Total number of students who read Biology - Number of students who read both Mathematics and Biology
= 25% - 15%
= 10%
Step 3: Calculate the probability that a student reads Mathematics given that he reads Biology.
We can use Bayes' theorem to calculate this probability.
P(Mathematics | Biology) = P(Biology | Mathematics) * P(Mathematics) / P(Biology)
- P(Mathematics) = 40%
- P(Biology) = 25%
- P(Biology | Mathematics) = The probability that a student reads Biology given that he reads Mathematics. This can be calculated as follows:
P(Biology | Mathematics) = (Number of students who read both Mathematics and Biology) / (Total number of students who read Mathematics)
= 15% / 40%
= 3/8
Substituting these values into Bayes' theorem, we get:
P(Mathematics | Biology) = (3/8 * 40%) / 25%
= 3/5
Therefore, the probability that a student reads Mathematics given that he reads Biology is 3/5 or option B.
- 40% of students read Mathematics.
- 25% of students read Biology.
- 15% of students read both Mathematics and Biology.
To find:
- The probability that a student reads Mathematics if it is known that he reads Biology.
Solution:
Step 1: Draw a Venn diagram representing the given information.
The rectangle represents the total number of students in the class. The circle on the left represents the number of students who read Mathematics, and the circle on the right represents the number of students who read Biology. The overlapping region represents the number of students who read both Mathematics and Biology.
Step 2: Calculate the number of students who read only Biology.
To do this, we subtract the number of students who read both Mathematics and Biology from the total number of students who read Biology.
Number of students who read only Biology = Total number of students who read Biology - Number of students who read both Mathematics and Biology
= 25% - 15%
= 10%
Step 3: Calculate the probability that a student reads Mathematics given that he reads Biology.
We can use Bayes' theorem to calculate this probability.
P(Mathematics | Biology) = P(Biology | Mathematics) * P(Mathematics) / P(Biology)
- P(Mathematics) = 40%
- P(Biology) = 25%
- P(Biology | Mathematics) = The probability that a student reads Biology given that he reads Mathematics. This can be calculated as follows:
P(Biology | Mathematics) = (Number of students who read both Mathematics and Biology) / (Total number of students who read Mathematics)
= 15% / 40%
= 3/8
Substituting these values into Bayes' theorem, we get:
P(Mathematics | Biology) = (3/8 * 40%) / 25%
= 3/5
Therefore, the probability that a student reads Mathematics given that he reads Biology is 3/5 or option B.
If an unbiased die is rolled once, the odds in favour of getting a point which is a multiple of 3 is- a)1:2
- b)2:1
- c)1:3
- d)3:1
Correct answer is option 'C'. Can you explain this answer?
If an unbiased die is rolled once, the odds in favour of getting a point which is a multiple of 3 is
a)
1:2
b)
2:1
c)
1:3
d)
3:1
|
|
Anu Kaur answered |
Total number =6
Getting a 'multiple of 3' = 2 . So, probability = 2/6 =1/3
Getting a 'multiple of 3' = 2 . So, probability = 2/6 =1/3
Two dice are thrown at a time. The probability that ‘the difference of nos shown is 1’ is- a)11/18
- b)5/18
- c)7/18
- d)none
Correct answer is option 'B'. Can you explain this answer?
Two dice are thrown at a time. The probability that ‘the difference of nos shown is 1’ is
a)
11/18
b)
5/18
c)
7/18
d)
none
|
Himanshi Tapkir answered |
Correct option is A)
Possible outcomes on rolling two dices are 36
Out of the possible outcomes, chances of getting numbers with difference one are as follows : 2,1;3,2;4,3;5,4;6,5 and vice versa in each case.
⇒ Number of favourable outcomes =10
Probability = Favourable Outcomes ÷ Total Outcomes =3610=185
Possible outcomes on rolling two dices are 36
Out of the possible outcomes, chances of getting numbers with difference one are as follows : 2,1;3,2;4,3;5,4;6,5 and vice versa in each case.
⇒ Number of favourable outcomes =10
Probability = Favourable Outcomes ÷ Total Outcomes =3610=185
A and B are two events such that P(A)= 1/3, P(B) = ¼, P(A+B)= 1/2, than P(B/A) is equal to- a)¼
- b)1/3
- c)1/2
- d)none
Correct answer is option 'A'. Can you explain this answer?
A and B are two events such that P(A)= 1/3, P(B) = ¼, P(A+B)= 1/2, than P(B/A) is equal to
a)
¼
b)
1/3
c)
1/2
d)
none
|
|
Mihir Banerjee answered |
Given, P(A) = 1/3, P(B), P(A ∩ B) = 1/2
We know that the conditional probability of B given A is given by:
P(B/A) = P(A ∩ B) / P(A)
Substituting the given values, we get:
P(B/A) = (1/2) / (1/3) = (1/2) x (3/1) = 3/2
But the probability of an event cannot be greater than 1. Therefore, option A is incorrect.
So, the correct answer is option D (none).
Explanation: It is not possible to calculate P(B/A) with the given information because we do not know the value of P(B ∩ A). We only know the value of P(A ∩ B) and not P(B ∩ A). Hence, none of the options given are correct.
We know that the conditional probability of B given A is given by:
P(B/A) = P(A ∩ B) / P(A)
Substituting the given values, we get:
P(B/A) = (1/2) / (1/3) = (1/2) x (3/1) = 3/2
But the probability of an event cannot be greater than 1. Therefore, option A is incorrect.
So, the correct answer is option D (none).
Explanation: It is not possible to calculate P(B/A) with the given information because we do not know the value of P(B ∩ A). We only know the value of P(A ∩ B) and not P(B ∩ A). Hence, none of the options given are correct.
Given that for two events A and B, P(A)=3/5, P(B)=2/3 and P(A union B) =3/4, what is P(A|B)?- a)0.655
- b)13/60
- c)31/60
- d)0.775
Correct answer is option 'D'. Can you explain this answer?
Given that for two events A and B, P(A)=3/5, P(B)=2/3 and P(A union B) =3/4, what is P(A|B)?
a)
0.655
b)
13/60
c)
31/60
d)
0.775
|
|
Freedom Institute answered |
P(A|B) = P(A&B)/P(B)
So first we have to find P(A&B)
P(AUB) = P(A) + P(B) - P(A&B)3/4 = 3/5 + 2/3 - P(A&B)
Multiply through by 60
45 = 36 + 40 - 60P(A&B)
45 = 76 - 60P(A&B)
-31 = -60P(A&B)
31/60 = P(A&B)
Go back to
P(A|B) = P(A&B)/P(B)
P(A|B) = (31/60)/(2/3)
P(A|B) = (31/60)(3/2)
P(A|B) = 31/40 = 0.775
So first we have to find P(A&B)
P(AUB) = P(A) + P(B) - P(A&B)3/4 = 3/5 + 2/3 - P(A&B)
Multiply through by 60
45 = 36 + 40 - 60P(A&B)
45 = 76 - 60P(A&B)
-31 = -60P(A&B)
31/60 = P(A&B)
Go back to
P(A|B) = P(A&B)/P(B)
P(A|B) = (31/60)/(2/3)
P(A|B) = (31/60)(3/2)
P(A|B) = 31/40 = 0.775
When unbiased coins are tossed. The probability of getting both heads or both tails is- a)½
- b)¾
- c)¼
- d)none
Correct answer is option 'A'. Can you explain this answer?
When unbiased coins are tossed. The probability of getting both heads or both tails is
a)
½
b)
¾
c)
¼
d)
none
|
Sai Karthik answered |
The 4 possibilities when 2 coins are tossed is TT,HH,TH and HT
the probability of getting two heads and two tails is
=favourable outcomes /total outcomes
=2/4
=1/2
the probability of getting two heads and two tails is
=favourable outcomes /total outcomes
=2/4
=1/2
Arun & Tarun appear for an interview for two vacancies. The probability of Arun's selection is 1/3 and that of Tarun's selection is 1/5. Find the probability that only one of them will be selected.
- a)2/5
- b)4/5
- c)6/5
- d)8/5
Correct answer is option 'A'. Can you explain this answer?
Arun & Tarun appear for an interview for two vacancies. The probability of Arun's selection is 1/3 and that of Tarun's selection is 1/5. Find the probability that only one of them will be selected.
a)
2/5
b)
4/5
c)
6/5
d)
8/5
|
|
Pallabi Khanna answered |
Can you please provide more context or information about who Arun is?
Two broad divisions of probability are- a)Subjective probability and objective probability
- b)Deductive probability and non–deductive probability
- c)Statistical probability and Mathematical probability
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
Two broad divisions of probability are
a)
Subjective probability and objective probability
b)
Deductive probability and non–deductive probability
c)
Statistical probability and Mathematical probability
d)
None of these.
|
|
Gaurav Chatterjee answered |
Divisions of Probability
Probability refers to the measure of the likelihood or chance of an event occurring. It is categorized into two broad divisions:
1. Subjective Probability
Subjective probability refers to the measure of probability based on personal judgement or experience. It is based on personal opinions and cannot be measured objectively. It relies on the individual's beliefs, biases, and intuition. It is often used in situations where there is a lack of data or information.
2. Objective Probability
Objective probability refers to the measure of probability based on data and evidence. It is based on empirical evidence and can be measured objectively. It relies on mathematical computations and statistical analysis. It is often used in situations where there is a large amount of data or information available.
Conclusion
In conclusion, probability is divided into subjective and objective probability. Subjective probability is based on personal judgement and experience while objective probability is based on data and evidence.
Probability refers to the measure of the likelihood or chance of an event occurring. It is categorized into two broad divisions:
1. Subjective Probability
Subjective probability refers to the measure of probability based on personal judgement or experience. It is based on personal opinions and cannot be measured objectively. It relies on the individual's beliefs, biases, and intuition. It is often used in situations where there is a lack of data or information.
2. Objective Probability
Objective probability refers to the measure of probability based on data and evidence. It is based on empirical evidence and can be measured objectively. It relies on mathematical computations and statistical analysis. It is often used in situations where there is a large amount of data or information available.
Conclusion
In conclusion, probability is divided into subjective and objective probability. Subjective probability is based on personal judgement and experience while objective probability is based on data and evidence.
The Theorem of Compound Probability states that for any two events A and B.- a)P(A ∩ B) = P(A) × P(B/A)
- b)P(A ∪ B) = P(A) × P(B/A)
- c)P(A ∪ B) = P(B) + P(B) – P(A ∩ B)
- d)P(A ∩ B) = P(A) × P(B)
Correct answer is option 'C'. Can you explain this answer?
The Theorem of Compound Probability states that for any two events A and B.
a)
P(A ∩ B) = P(A) × P(B/A)
b)
P(A ∪ B) = P(A) × P(B/A)
c)
P(A ∪ B) = P(B) + P(B) – P(A ∩ B)
d)
P(A ∩ B) = P(A) × P(B)
|
Sai Joshi answered |
Correct Answer :- C
Explanation : If two events, A and B, are mutually exclusive, then the probability that either A or B occurs is the sum of their probabilities.
For mutually inclusive events, P (A or B) = P(A) + P(B) - P(A and B).
There are three boxes with the following composition:Box I: 5 Red + 7 White + 6 Blue ballsBox II: 4 Red + 8 White + 6 Blue ballsBox III: 3 Red + 4 White + 2 Blue ballsIf one ball is drawn at random, then what is the probability that they would be of same colour?- a)89/729
- b)97/729
- c)82/729
- d)23/32
Correct answer is option 'A'. Can you explain this answer?
There are three boxes with the following composition:
Box I: 5 Red + 7 White + 6 Blue balls
Box II: 4 Red + 8 White + 6 Blue balls
Box III: 3 Red + 4 White + 2 Blue balls
If one ball is drawn at random, then what is the probability that they would be of same colour?
a)
89/729
b)
97/729
c)
82/729
d)
23/32
|
|
Ruchi Mishra answered |
Solution:
To find the probability that the ball drawn is of the same color, we need to find the probability of drawing a red ball or a white ball or a blue ball from any of the three boxes.
Probability of drawing a red ball:
Total red balls = 5 + 4 + 3 = 12
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a red ball = 12/45 = 4/15
Probability of drawing a white ball:
Total white balls = 7 + 8 + 4 = 19
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a white ball = 19/45
Probability of drawing a blue ball:
Total blue balls = 6 + 6 + 2 = 14
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a blue ball = 14/45
Now, we need to find the probability of drawing a ball of the same color, which can be calculated as follows:
Probability of drawing a ball of the same color = Probability of drawing a red ball + Probability of drawing a white ball + Probability of drawing a blue ball
Probability of drawing a ball of the same color = 4/15 + 19/45 + 14/45
Probability of drawing a ball of the same color = 89/225
Simplifying the fraction, we get:
Probability of drawing a ball of the same color = 89/729
Therefore, the correct option is (a) 89/729.
To find the probability that the ball drawn is of the same color, we need to find the probability of drawing a red ball or a white ball or a blue ball from any of the three boxes.
Probability of drawing a red ball:
Total red balls = 5 + 4 + 3 = 12
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a red ball = 12/45 = 4/15
Probability of drawing a white ball:
Total white balls = 7 + 8 + 4 = 19
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a white ball = 19/45
Probability of drawing a blue ball:
Total blue balls = 6 + 6 + 2 = 14
Total number of balls = 5 + 7 + 6 + 4 + 8 + 6 + 3 + 4 + 2 = 45
Therefore, the probability of drawing a blue ball = 14/45
Now, we need to find the probability of drawing a ball of the same color, which can be calculated as follows:
Probability of drawing a ball of the same color = Probability of drawing a red ball + Probability of drawing a white ball + Probability of drawing a blue ball
Probability of drawing a ball of the same color = 4/15 + 19/45 + 14/45
Probability of drawing a ball of the same color = 89/225
Simplifying the fraction, we get:
Probability of drawing a ball of the same color = 89/729
Therefore, the correct option is (a) 89/729.
The probability of Girl getting scholarship is 0.6 and the same probability for Boy is 0.8. Find the probability that at least one of the categories getiing scholarship.
- a)0.32
- b)0.44
- c)0.52
- d)None of the above.
Correct answer is option 'C'. Can you explain this answer?
The probability of Girl getting scholarship is 0.6 and the same probability for Boy is 0.8. Find the probability that at least one of the categories getiing scholarship.
a)
0.32
b)
0.44
c)
0.52
d)
None of the above.
|
|
Malavika Basak answered |
Solution:
Given, the probability of a girl getting a scholarship is 0.6 and the probability of a boy getting a scholarship is 0.8.
Let A be the event that a girl gets a scholarship and B be the event that a boy gets a scholarship.
We need to find the probability that at least one of the categories is getting a scholarship, i.e., P(A∪B).
Using the formula for the probability of the union of two events, we have:
P(A∪B) = P(A) + P(B) - P(A∩B)
where P(A∩B) is the probability that both a girl and a boy get a scholarship.
Since the events A and B are independent (i.e., the probability of one event does not affect the probability of the other event), we have:
P(A∩B) = P(A) × P(B) = 0.6 × 0.8 = 0.48
Substituting the values in the above equation, we get:
P(A∪B) = 0.6 + 0.8 - 0.48 = 0.92
Hence, the probability that at least one of the categories is getting a scholarship is 0.92.
Therefore, the correct answer is option 'C'.
Given, the probability of a girl getting a scholarship is 0.6 and the probability of a boy getting a scholarship is 0.8.
Let A be the event that a girl gets a scholarship and B be the event that a boy gets a scholarship.
We need to find the probability that at least one of the categories is getting a scholarship, i.e., P(A∪B).
Using the formula for the probability of the union of two events, we have:
P(A∪B) = P(A) + P(B) - P(A∩B)
where P(A∩B) is the probability that both a girl and a boy get a scholarship.
Since the events A and B are independent (i.e., the probability of one event does not affect the probability of the other event), we have:
P(A∩B) = P(A) × P(B) = 0.6 × 0.8 = 0.48
Substituting the values in the above equation, we get:
P(A∪B) = 0.6 + 0.8 - 0.48 = 0.92
Hence, the probability that at least one of the categories is getting a scholarship is 0.92.
Therefore, the correct answer is option 'C'.
Following are the wages of 8 workers in rupees:50, 62, 40, 70, 45, 56, 32, 45If one of the workers is selected at random, what is the probability that his wage would be lower than the average wage?- a)0.625
- b)0.500
- c)0.375
- d)0.450
Correct answer is option 'B'. Can you explain this answer?
Following are the wages of 8 workers in rupees:
50, 62, 40, 70, 45, 56, 32, 45
If one of the workers is selected at random, what is the probability that his wage would be lower than the average wage?
a)
0.625
b)
0.500
c)
0.375
d)
0.450
|
|
Vaishnavi Gupta answered |
Given:
Wages of 8 workers in rupees:50, 62, 40, 70, 45, 56, 32, 45
To find:
Probability that a worker's wage would be lower than the average wage.
Solution:
Step 1: Calculate the average wage
Average wage = (50+62+40+70+45+56+32+45)/8
Average wage = 47.5
Step 2: Find the number of workers whose wage is lower than the average wage
Number of workers whose wage is lower than the average wage = 3 (out of 8)
Step 3: Calculate the probability
Probability = Number of workers whose wage is lower than the average wage / Total number of workers
Probability = 3/8
Probability = 0.5
Therefore, the probability that a worker's wage would be lower than the average wage is 0.5 or 50%. Hence, option B is the correct answer.
Wages of 8 workers in rupees:50, 62, 40, 70, 45, 56, 32, 45
To find:
Probability that a worker's wage would be lower than the average wage.
Solution:
Step 1: Calculate the average wage
Average wage = (50+62+40+70+45+56+32+45)/8
Average wage = 47.5
Step 2: Find the number of workers whose wage is lower than the average wage
Number of workers whose wage is lower than the average wage = 3 (out of 8)
Step 3: Calculate the probability
Probability = Number of workers whose wage is lower than the average wage / Total number of workers
Probability = 3/8
Probability = 0.5
Therefore, the probability that a worker's wage would be lower than the average wage is 0.5 or 50%. Hence, option B is the correct answer.
(Direction 16 - 38) Write down the correct answers. Each question carries 2 marks.Q.Two balls are drawn from a bag containing 5 white and 7 black balls at random. What is the probability that they would be of different colours? - a)35/66
- b)30/66
- c)12/66
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
(Direction 16 - 38) Write down the correct answers. Each question carries 2 marks.
Q.Two balls are drawn from a bag containing 5 white and 7 black balls at random. What is the probability that they would be of different colours?
a)
35/66
b)
30/66
c)
12/66
d)
None of these
|
|
Deepika Nambiar answered |
Solution:
Given, a bag contains 5 white and 7 black balls.
We need to find the probability of drawing two balls of different colours.
Total number of ways to draw two balls out of 12 balls = 12C2 = (12 × 11) / (2 × 1) = 66
Number of ways to draw two balls of different colours = Number of ways to draw one white and one black ball + Number of ways to draw one black and one white ball
= (5C1 × 7C1) + (7C1 × 5C1)
= (5 × 7) + (7 × 5)
= 35 + 35
= 70
Therefore, the probability of drawing two balls of different colours = Number of ways to draw two balls of different colours / Total number of ways to draw two balls
= 70 / 66
= 35 / 33
Hence, the correct option is (a) 35/66.
Given, a bag contains 5 white and 7 black balls.
We need to find the probability of drawing two balls of different colours.
Total number of ways to draw two balls out of 12 balls = 12C2 = (12 × 11) / (2 × 1) = 66
Number of ways to draw two balls of different colours = Number of ways to draw one white and one black ball + Number of ways to draw one black and one white ball
= (5C1 × 7C1) + (7C1 × 5C1)
= (5 × 7) + (7 × 5)
= 35 + 35
= 70
Therefore, the probability of drawing two balls of different colours = Number of ways to draw two balls of different colours / Total number of ways to draw two balls
= 70 / 66
= 35 / 33
Hence, the correct option is (a) 35/66.
In how many ways a team of 5 can be made out of 7 Boys and 8 Girls, if 2 Girls are compulsory to form a Team.- a)2,646
- b)1,722
- c)2,702
- d)980
Correct answer is option 'C'. Can you explain this answer?
In how many ways a team of 5 can be made out of 7 Boys and 8 Girls, if 2 Girls are compulsory to form a Team.
a)
2,646
b)
1,722
c)
2,702
d)
980
|
|
Anu Sen answered |
Solution:
In this problem, we have to form a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory.
Step 1: Select 2 girls out of 8 girls in 8C2 ways. (Since 2 girls are compulsory)
Step 2: Select 3 members from remaining 5 girls and 7 boys in (5C3 + 7C3) ways.
Step 3: Total number of ways of forming a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory = 8C2 × (5C3 + 7C3)
Step 4: Calculation
8C2 = (8!)/(2! × 6!) = 28
5C3 = (5!)/(3! × 2!) = 10
7C3 = (7!)/(3! × 4!) = 35
Total number of ways of forming a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory = 28 × (10+35) = 28 × 45 = 1260
Hence, option (c) 2,702 is the correct answer.
In this problem, we have to form a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory.
Step 1: Select 2 girls out of 8 girls in 8C2 ways. (Since 2 girls are compulsory)
Step 2: Select 3 members from remaining 5 girls and 7 boys in (5C3 + 7C3) ways.
Step 3: Total number of ways of forming a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory = 8C2 × (5C3 + 7C3)
Step 4: Calculation
8C2 = (8!)/(2! × 6!) = 28
5C3 = (5!)/(3! × 2!) = 10
7C3 = (7!)/(3! × 4!) = 35
Total number of ways of forming a team of 5 members out of 7 boys and 8 girls, where 2 girls are compulsory = 28 × (10+35) = 28 × 45 = 1260
Hence, option (c) 2,702 is the correct answer.
Values of a random variable are- a)always positive numbers.
- b)always positive real numbers.
- c)real numbers.
- d)natural numbers
Correct answer is option 'C'. Can you explain this answer?
Values of a random variable are
a)
always positive numbers.
b)
always positive real numbers.
c)
real numbers.
d)
natural numbers
|
|
Shanaya Dasgupta answered |
Explanation:
Random variable is a variable whose value is subject to variations due to chance, i.e., it is a variable whose value is determined by the outcome of a random event. The values of a random variable can be classified into four categories:
a) Always positive numbers: This is not always true for all random variables. For example, the outcome of a coin toss can be either heads or tails, which are not positive numbers.
b) Always positive real numbers: This is also not always true for all random variables. For example, the outcome of rolling a six-sided die can be any integer from 1 to 6, which are not necessarily positive real numbers.
c) Real numbers: This is the correct answer. The values of a random variable can be any real number. For example, the height of a person, the weight of an object, or the temperature in a room can all be random variables that take on real number values.
d) Natural numbers: This is not always true for all random variables. For example, the outcome of rolling a six-sided die can be any integer from 1 to 6, which are natural numbers, but the outcome of flipping a coin cannot be a natural number.
In conclusion, the values of a random variable can be any real number, which is option 'C'.
Random variable is a variable whose value is subject to variations due to chance, i.e., it is a variable whose value is determined by the outcome of a random event. The values of a random variable can be classified into four categories:
a) Always positive numbers: This is not always true for all random variables. For example, the outcome of a coin toss can be either heads or tails, which are not positive numbers.
b) Always positive real numbers: This is also not always true for all random variables. For example, the outcome of rolling a six-sided die can be any integer from 1 to 6, which are not necessarily positive real numbers.
c) Real numbers: This is the correct answer. The values of a random variable can be any real number. For example, the height of a person, the weight of an object, or the temperature in a room can all be random variables that take on real number values.
d) Natural numbers: This is not always true for all random variables. For example, the outcome of rolling a six-sided die can be any integer from 1 to 6, which are natural numbers, but the outcome of flipping a coin cannot be a natural number.
In conclusion, the values of a random variable can be any real number, which is option 'C'.
Two unbiased coins are tossed. The probability of obtaining one head and one tail is
- a)¼
- b)2/4
- c)¾
- d)none
Correct answer is option 'B'. Can you explain this answer?
Two unbiased coins are tossed. The probability of obtaining one head and one tail is
a)
¼
b)
2/4
c)
¾
d)
none
|
|
Jyoti Nair answered |
Explanation:
When two unbiased coins are tossed, there are four possible outcomes: HH, HT, TH, and TT. Out of these, only two outcomes have one head and one tail: HT and TH.
Therefore, the probability of obtaining one head and one tail is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Solution:
Number of favorable outcomes: 2 (HT and TH)
Total number of possible outcomes: 4 (HH, HT, TH, and TT)
Probability of obtaining one head and one tail:
= Number of favorable outcomes / Total number of possible outcomes
= 2 / 4
= 1/2
Hence, the correct answer is option B: 2/4.
When two unbiased coins are tossed, there are four possible outcomes: HH, HT, TH, and TT. Out of these, only two outcomes have one head and one tail: HT and TH.
Therefore, the probability of obtaining one head and one tail is the ratio of the number of favorable outcomes to the total number of possible outcomes.
Solution:
Number of favorable outcomes: 2 (HT and TH)
Total number of possible outcomes: 4 (HH, HT, TH, and TT)
Probability of obtaining one head and one tail:
= Number of favorable outcomes / Total number of possible outcomes
= 2 / 4
= 1/2
Hence, the correct answer is option B: 2/4.
Two unbiased coins are tossed. The probability of obtaining at least one head is- a)¼
- b)2/4
- c)¾
- d)none
Correct answer is option 'C'. Can you explain this answer?
Two unbiased coins are tossed. The probability of obtaining at least one head is
a)
¼
b)
2/4
c)
¾
d)
none
|
Pathan Mahenoor answered |
Total outcomes = (HH,HT,TT,TH) = 4
atleast one head means 1 head or more
p(HT,TH,HH) = 3/4
atleast one head means 1 head or more
p(HT,TH,HH) = 3/4
(Direction 1 - 40) Write down the correct answers. Each question carRies 1 mark.Q. Initially, probability was a branch of- a)Physics
- b)Statistics
- c)Mathematics
- d)Economics.
Correct answer is option 'C'. Can you explain this answer?
(Direction 1 - 40) Write down the correct answers. Each question carRies 1 mark.
Q. Initially, probability was a branch of
a)
Physics
b)
Statistics
c)
Mathematics
d)
Economics.
|
|
Freedom Institute answered |
Probability theory is the branch of Mathematics concerned with analysis of random phenomena.
If all the values taken by a random variable are equal then- a)its expected value is zero
- b)its standard deviation is zero
- c)its standard deviation is positive
- d)its standard deviation is a real number
Correct answer is option 'B'. Can you explain this answer?
If all the values taken by a random variable are equal then
a)
its expected value is zero
b)
its standard deviation is zero
c)
its standard deviation is positive
d)
its standard deviation is a real number
|
|
Pranav Gupta answered |
Explanation:
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. If all the values taken by a random variable are equal, then the variable is a constant random variable.
a) its expected value is zero: Since all the values are equal, the expected value will be the same as those values. Therefore, the expected value will be the constant value itself, which is not zero in general.
b) its standard deviation is zero: The standard deviation of a random variable is a measure of the spread of its values. Since all the values are equal, there is no spread, and the standard deviation is zero.
c) its standard deviation is positive: This option is incorrect since the standard deviation cannot be positive if all the values are equal.
d) its standard deviation is a real number: This option is incorrect since every real number is a real number, and it does not provide any useful information.
Therefore, the correct answer is option B.
A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. If all the values taken by a random variable are equal, then the variable is a constant random variable.
a) its expected value is zero: Since all the values are equal, the expected value will be the same as those values. Therefore, the expected value will be the constant value itself, which is not zero in general.
b) its standard deviation is zero: The standard deviation of a random variable is a measure of the spread of its values. Since all the values are equal, there is no spread, and the standard deviation is zero.
c) its standard deviation is positive: This option is incorrect since the standard deviation cannot be positive if all the values are equal.
d) its standard deviation is a real number: This option is incorrect since every real number is a real number, and it does not provide any useful information.
Therefore, the correct answer is option B.
If p : q are the odds in favour of an event, then the probability of that event is- a)p/q
- b)p/p+q
- c)q/p+q
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If p : q are the odds in favour of an event, then the probability of that event is
a)
p/q
b)
p/p+q
c)
q/p+q
d)
none of these
|
Rajat Patel answered |
Explanation:
Total number of cases=p+q
Favourable cases=p
Probability of that event is=p/(p+q)
Two coins are tossed simultaneously. What is the probability that the second coin would show a tail given that the first coin has shown a head?
- a)0.25
- b)0.50
- c)0.75
- d)0.125
Correct answer is option 'A'. Can you explain this answer?
Two coins are tossed simultaneously. What is the probability that the second coin would show a tail given that the first coin has shown a head?
a)
0.25
b)
0.50
c)
0.75
d)
0.125
|
|
Freedom Institute answered |
The sample space for two coins is {HH,HT,TH,TT}, n(S)=4
Favourable outcomes for the first coin head and second coin showing tail are HT, n(E)=1
Hence, the probability that the second coin would show a tail given that the first coin has shown a head is n(E)/n(S) = 1/4 = 0.25
If A denotes that a student reads in a school and B denotes that he plays cricket, then- a)P(A ∩ B) = 1
- b)P(A ∪ B) = 1
- c)P(A ∩ B) = 0
- d)P(A) = P(B)
Correct answer is option 'C'. Can you explain this answer?
If A denotes that a student reads in a school and B denotes that he plays cricket, then
a)
P(A ∩ B) = 1
b)
P(A ∪ B) = 1
c)
P(A ∩ B) = 0
d)
P(A) = P(B)
|
|
Hrishikesh Mukherjee answered |
Probability of a student reading in school and playing cricket
To solve this problem, we need to understand the definitions of probability and set theory.
Probability: Probability is the measure of the likelihood of an event occurring.
Set theory: Set theory is the branch of mathematics that deals with the study of sets, which are collections of objects.
In this problem, we are given two events:
A: A student reads in a school
B: A student plays cricket
We need to find the probability of the event A and B occurring together.
Intersection of two events
The intersection of two events is the set of elements that belong to both events.
In set theory, the intersection of two sets A and B is denoted by A ∩ B.
For example, if A = {1, 2, 3} and B = {2, 3, 4}, then A ∩ B = {2, 3}.
Probability of intersection of two events
The probability of the intersection of two events A and B is denoted by P(A ∩ B).
P(A ∩ B) represents the probability that both A and B occur.
Formula:
P(A ∩ B) = P(A) * P(B|A)
where P(B|A) represents the probability of B occurring given that A has already occurred.
If A and B are independent events, then P(B|A) = P(B), and the formula reduces to:
P(A ∩ B) = P(A) * P(B)
Solution
In this problem, we are not given any information about the relationship between A and B.
Therefore, we cannot assume that they are independent events.
Since we do not have any information about the relationship between A and B, we cannot calculate the probability of their intersection.
Therefore, the correct answer is option C: P(A ∩ B) = 0.
To solve this problem, we need to understand the definitions of probability and set theory.
Probability: Probability is the measure of the likelihood of an event occurring.
Set theory: Set theory is the branch of mathematics that deals with the study of sets, which are collections of objects.
In this problem, we are given two events:
A: A student reads in a school
B: A student plays cricket
We need to find the probability of the event A and B occurring together.
Intersection of two events
The intersection of two events is the set of elements that belong to both events.
In set theory, the intersection of two sets A and B is denoted by A ∩ B.
For example, if A = {1, 2, 3} and B = {2, 3, 4}, then A ∩ B = {2, 3}.
Probability of intersection of two events
The probability of the intersection of two events A and B is denoted by P(A ∩ B).
P(A ∩ B) represents the probability that both A and B occur.
Formula:
P(A ∩ B) = P(A) * P(B|A)
where P(B|A) represents the probability of B occurring given that A has already occurred.
If A and B are independent events, then P(B|A) = P(B), and the formula reduces to:
P(A ∩ B) = P(A) * P(B)
Solution
In this problem, we are not given any information about the relationship between A and B.
Therefore, we cannot assume that they are independent events.
Since we do not have any information about the relationship between A and B, we cannot calculate the probability of their intersection.
Therefore, the correct answer is option C: P(A ∩ B) = 0.
Let P be a probability function on S = {X1 , X2 , X3} if P(X1)= ¼ and P(X3) = 1/3 then P (X2) is equal to- a)5/12
- b)7/12
- c)3/4
- d)none
Correct answer is option 'A'. Can you explain this answer?
Let P be a probability function on S = {X1 , X2 , X3} if P(X1)= ¼ and P(X3) = 1/3 then P (X2) is equal to
a)
5/12
b)
7/12
c)
3/4
d)
none
|
|
Muskaan Tiwari answered |
Given information:
- S = {X1, X2, X3}
- P(X1) =
- P(X3) = 1/3
To find:
- P(X2)
Solution:
- As P is a probability function, the sum of probabilities of all elements in the sample space S must be 1.
- Therefore, P(X1) + P(X2) + P(X3) = 1
- Using the given values, we get:
+ + 1/3 = 1
+ = 2/3
- Therefore, P(X2) =
Answer:
- P(X2) =
- Option A, 5/12, is the correct answer.
- S = {X1, X2, X3}
- P(X1) =
- P(X3) = 1/3
To find:
- P(X2)
Solution:
- As P is a probability function, the sum of probabilities of all elements in the sample space S must be 1.
- Therefore, P(X1) + P(X2) + P(X3) = 1
- Using the given values, we get:
+ + 1/3 = 1
+ = 2/3
- Therefore, P(X2) =
Answer:
- P(X2) =
- Option A, 5/12, is the correct answer.
If events A and B are independent and P(A)= 2/3 , P(B)= 3/5 then P(A+B)is equal to- a)13/15
- b)6/15
- c)1/15
- d)none
Correct answer is option 'A'. Can you explain this answer?
If events A and B are independent and P(A)= 2/3 , P(B)= 3/5 then P(A+B)is equal to
a)
13/15
b)
6/15
c)
1/15
d)
none
|
|
Deepika Desai answered |
Given:
- Events A and B are independent
- P(A)= 2/3
- P(B)= 3/5
To find:
- P(A ∩ B)
Approach:
- Use the formula for the probability of the intersection of independent events
- Substitute the given values
- Simplify the expression
Formula:
- P(A ∩ B) = P(A) * P(B)
Calculation:
- P(A ∩ B) = P(A) * P(B)
- P(A ∩ B) = (2/3) * (3/5)
- P(A ∩ B) = 6/15
- P(A ∩ B) = 2/5
Answer:
- The correct option is A) 13/15.
- Events A and B are independent
- P(A)= 2/3
- P(B)= 3/5
To find:
- P(A ∩ B)
Approach:
- Use the formula for the probability of the intersection of independent events
- Substitute the given values
- Simplify the expression
Formula:
- P(A ∩ B) = P(A) * P(B)
Calculation:
- P(A ∩ B) = P(A) * P(B)
- P(A ∩ B) = (2/3) * (3/5)
- P(A ∩ B) = 6/15
- P(A ∩ B) = 2/5
Answer:
- The correct option is A) 13/15.
The probability of selecting sample of size 'n' out of a population of size N by simple random sampling with replacement is:- a) 1/N
- b)1/Nn
- c)1/Ncn
- d)
Correct answer is option 'B'. Can you explain this answer?
The probability of selecting sample of size 'n' out of a population of size N by simple random sampling with replacement is:
a)
1/N
b)
1/Nn
c)
1/Ncn
d)
|
Mohit Choudhary answered |
Total sample space = N ^ n
Event =1
Probability = (Event)/(sample space) = 1/( N^ n)
A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at random. What is the probability that none of the balls drawn is blue?- a)10/21
- b)11/21
- c)2/7
- d)5/7
Correct answer is option 'A'. Can you explain this answer?
A box contains 2 red, 3 green and 2 blue balls. Two bails are drawn at random. What is the probability that none of the balls drawn is blue?
a)
10/21
b)
11/21
c)
2/7
d)
5/7
|
Aman Chaudhary answered |
Given:
- Box contains 2 red, 3 green and 2 blue balls.
- Two balls are drawn at random.
To find:
- Probability that none of the balls drawn is blue.
Solution:
Total number of ways to draw 2 balls out of 7 balls = 7C2 = 21
Let A be the event of drawing a non-blue ball in the first draw and B be the event of drawing a non-blue ball in the second draw.
P(A) = (2 red balls + 3 green balls) / 7 balls = 5/7
P(B | A) = probability of drawing a non-blue ball in the second draw given that a non-blue ball was drawn in the first draw
= (1 red ball + 2 green balls) / 6 balls = 3/6 = 1/2
P(A and B) = P(A) x P(B | A)
= (5/7) x (1/2)
= 5/14
Therefore, the probability of drawing two non-blue balls = P(A and B) = 5/14
The probability of not drawing any blue ball = 1 - probability of drawing at least one blue ball
= 1 - (number of ways to draw 1 blue ball x number of ways to draw 1 non-blue ball) / total number of ways to draw 2 balls
= 1 - (2C1 x 5C1) / 21
= 1 - (10/21)
= 11/21
Answer: The probability that none of the balls drawn is blue is 11/21, which is not among the given options. Therefore, the given answer key must be incorrect.
- Box contains 2 red, 3 green and 2 blue balls.
- Two balls are drawn at random.
To find:
- Probability that none of the balls drawn is blue.
Solution:
Total number of ways to draw 2 balls out of 7 balls = 7C2 = 21
Let A be the event of drawing a non-blue ball in the first draw and B be the event of drawing a non-blue ball in the second draw.
P(A) = (2 red balls + 3 green balls) / 7 balls = 5/7
P(B | A) = probability of drawing a non-blue ball in the second draw given that a non-blue ball was drawn in the first draw
= (1 red ball + 2 green balls) / 6 balls = 3/6 = 1/2
P(A and B) = P(A) x P(B | A)
= (5/7) x (1/2)
= 5/14
Therefore, the probability of drawing two non-blue balls = P(A and B) = 5/14
The probability of not drawing any blue ball = 1 - probability of drawing at least one blue ball
= 1 - (number of ways to draw 1 blue ball x number of ways to draw 1 non-blue ball) / total number of ways to draw 2 balls
= 1 - (2C1 x 5C1) / 21
= 1 - (10/21)
= 11/21
Answer: The probability that none of the balls drawn is blue is 11/21, which is not among the given options. Therefore, the given answer key must be incorrect.
In a class 40 % students read Mathematics, 25 % Biology and 15 % both Mathematics and Biology. One student is select at random.The probability that he reads Biology if he reads Mathematics- a)7/8
- b)1/8
- c)3/8
- d)none
Correct answer is option 'C'. Can you explain this answer?
In a class 40 % students read Mathematics, 25 % Biology and 15 % both Mathematics and Biology. One student is select at random.The probability that he reads Biology if he reads Mathematics
a)
7/8
b)
1/8
c)
3/8
d)
none
|
|
Aditya Das answered |
Solution:
Given,
percentage of students who read Mathematics = 40%
percentage of students who read Biology = 25%
percentage of students who read both Mathematics and Biology = 15%
Let,
M = event that a student reads Mathematics
B = event that a student reads Biology
To find: Probability of reading Biology given that the student reads Mathematics i.e. P(B|M)
We know that,
P(B ∪ M) = P(B) + P(M) - P(B ∩ M) (where ∪ represents union and ∩ represents intersection)
Also,
P(B ∪ M) = 100% (as all students either read Mathematics or Biology or both)
Therefore,
P(B) + P(M) - P(B ∩ M) = 100%
25% + 40% - 15% = 50%
Now, we can use the formula of conditional probability to find P(B|M):
P(B|M) = P(B ∩ M) / P(M)
= 15% / 40%
= 3/8
Hence, the correct option is (c) 3/8.
Given,
percentage of students who read Mathematics = 40%
percentage of students who read Biology = 25%
percentage of students who read both Mathematics and Biology = 15%
Let,
M = event that a student reads Mathematics
B = event that a student reads Biology
To find: Probability of reading Biology given that the student reads Mathematics i.e. P(B|M)
We know that,
P(B ∪ M) = P(B) + P(M) - P(B ∩ M) (where ∪ represents union and ∩ represents intersection)
Also,
P(B ∪ M) = 100% (as all students either read Mathematics or Biology or both)
Therefore,
P(B) + P(M) - P(B ∩ M) = 100%
25% + 40% - 15% = 50%
Now, we can use the formula of conditional probability to find P(B|M):
P(B|M) = P(B ∩ M) / P(M)
= 15% / 40%
= 3/8
Hence, the correct option is (c) 3/8.
When unbiased coins are tossed. The probability of obtaining not more than 3 heads is- a)¾
- b)½
- c)1
- d)0
Correct answer is option 'C'. Can you explain this answer?
When unbiased coins are tossed. The probability of obtaining not more than 3 heads is
a)
¾
b)
½
c)
1
d)
0
|
|
Maheshwar Sharma answered |
Solution:
Given, unbiased coins are tossed.
We need to find the probability of obtaining not more than 3 heads.
Let us consider the possible outcomes when 4 coins are tossed:
- All tails: TTTT
- One head: HTTT, THTT, TTHT, TTTH
- Two heads: HHTT, HTHT, HTTH, THHT, THTH, TTHH
- Three heads: HHHT, HHTH, HTHH, THHH
- All heads: HHHH
Out of these possible outcomes, there are 4 outcomes where we obtain not more than 3 heads.
Therefore, the probability of obtaining not more than 3 heads = 4/16 = 1/4 = 0.25.
Hence, the correct answer is option 'C'.
Note: The probability of obtaining not more than 3 heads can also be calculated using the binomial distribution formula as follows:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
where X is the number of heads obtained in 4 tosses of a coin.
Using the binomial distribution formula, we get:
P(X ≤ 3) = (4C0 * (1/2)^0 * (1/2)^4) + (4C1 * (1/2)^1 * (1/2)^3) + (4C2 * (1/2)^2 * (1/2)^2) + (4C3 * (1/2)^3 * (1/2)^1)
= (1 * 1 * 1/16) + (4 * 1/2 * 1/8) + (6 * 1/4 * 1/4) + (4 * 1/8 * 1/2)
= 1/16 + 1/4 + 3/16 + 1/8
= 1/4
Therefore, the probability of obtaining not more than 3 heads = 1/4 = 0.25.
Given, unbiased coins are tossed.
We need to find the probability of obtaining not more than 3 heads.
Let us consider the possible outcomes when 4 coins are tossed:
- All tails: TTTT
- One head: HTTT, THTT, TTHT, TTTH
- Two heads: HHTT, HTHT, HTTH, THHT, THTH, TTHH
- Three heads: HHHT, HHTH, HTHH, THHH
- All heads: HHHH
Out of these possible outcomes, there are 4 outcomes where we obtain not more than 3 heads.
Therefore, the probability of obtaining not more than 3 heads = 4/16 = 1/4 = 0.25.
Hence, the correct answer is option 'C'.
Note: The probability of obtaining not more than 3 heads can also be calculated using the binomial distribution formula as follows:
P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
where X is the number of heads obtained in 4 tosses of a coin.
Using the binomial distribution formula, we get:
P(X ≤ 3) = (4C0 * (1/2)^0 * (1/2)^4) + (4C1 * (1/2)^1 * (1/2)^3) + (4C2 * (1/2)^2 * (1/2)^2) + (4C3 * (1/2)^3 * (1/2)^1)
= (1 * 1 * 1/16) + (4 * 1/2 * 1/8) + (6 * 1/4 * 1/4) + (4 * 1/8 * 1/2)
= 1/16 + 1/4 + 3/16 + 1/8
= 1/4
Therefore, the probability of obtaining not more than 3 heads = 1/4 = 0.25.
For two events A and B, P(A ∪ B) = P(A) + P(A) only when- a)A and B are equally likely events
- b)A and B are exhaustive events
- c)A and B are mutually independent
- d)A and B are mutually exclusive
Correct answer is option 'D'. Can you explain this answer?
For two events A and B, P(A ∪ B) = P(A) + P(A) only when
a)
A and B are equally likely events
b)
A and B are exhaustive events
c)
A and B are mutually independent
d)
A and B are mutually exclusive
|
|
Srsps answered |
Definition: Two events are mutually exclusive (or disjoint) if they cannot occur at the same time. In other words, the occurrence of one event excludes the possibility of the other event occurring.
Hence, the intersection of both the events must be zero.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
and as the intersection is 0 in mutually exclusive events
P(A ∪ B) = P(A) + P(B)
Hence, the intersection of both the events must be zero.
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
and as the intersection is 0 in mutually exclusive events
P(A ∪ B) = P(A) + P(B)
Two unbiased dice are thrown. The Expected value of the sum of numbers on the upper side is;- a)3.5
- b)7
- c)12
- d)6
Correct answer is option 'B'. Can you explain this answer?
Two unbiased dice are thrown. The Expected value of the sum of numbers on the upper side is;
a)
3.5
b)
7
c)
12
d)
6
|
|
Anuj Roy answered |
Solution:
When two dice are thrown, the possible outcomes can be represented by a sample space of 36 outcomes.
Let A be the event that the sum of the numbers on the upper side of the dice is k, where k = 2, 3, ..., 12.
Then, the probability of A is given by:
P(A) = number of outcomes in A / total number of outcomes
The number of outcomes in A can be determined by counting the number of ways that k can be obtained as the sum of two numbers on the dice. For example, if k = 7, there are six ways to obtain this sum:
1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1
Therefore, the number of outcomes in A is:
number of outcomes in A = 6 if k = 7
number of outcomes in A = 5 if k = 6 or 8
number of outcomes in A = 4 if k = 5 or 9
number of outcomes in A = 3 if k = 4 or 10
number of outcomes in A = 2 if k = 3 or 11
number of outcomes in A = 1 if k = 2 or 12
The total number of outcomes is 36, so the probabilities of the events A are:
P(A=2) = 1/36
P(A=3) = 2/36
P(A=4) = 3/36
P(A=5) = 4/36
P(A=6) = 5/36
P(A=7) = 6/36
P(A=8) = 5/36
P(A=9) = 4/36
P(A=10) = 3/36
P(A=11) = 2/36
P(A=12) = 1/36
The expected value of the sum of the numbers on the upper side of the dice is given by:
E(X) = Σk P(X=k)
where k is the sum of the numbers on the upper side of the dice.
Using the probabilities calculated above, we can find that:
E(X) = 2×1/36 + 3×2/36 + 4×3/36 + 5×4/36 + 6×5/36 + 7×6/36 + 8×5/36 + 9×4/36 + 10×3/36 + 11×2/36 + 12×1/36
Simplifying this expression, we get:
E(X) = (2+3+4+5+6+7+8+9+10+11+12)/2
E(X) = 7
Therefore, the expected value of the sum of the numbers on the upper side of the dice is 7.
Hence, option B is the correct answer.
When two dice are thrown, the possible outcomes can be represented by a sample space of 36 outcomes.
Let A be the event that the sum of the numbers on the upper side of the dice is k, where k = 2, 3, ..., 12.
Then, the probability of A is given by:
P(A) = number of outcomes in A / total number of outcomes
The number of outcomes in A can be determined by counting the number of ways that k can be obtained as the sum of two numbers on the dice. For example, if k = 7, there are six ways to obtain this sum:
1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1
Therefore, the number of outcomes in A is:
number of outcomes in A = 6 if k = 7
number of outcomes in A = 5 if k = 6 or 8
number of outcomes in A = 4 if k = 5 or 9
number of outcomes in A = 3 if k = 4 or 10
number of outcomes in A = 2 if k = 3 or 11
number of outcomes in A = 1 if k = 2 or 12
The total number of outcomes is 36, so the probabilities of the events A are:
P(A=2) = 1/36
P(A=3) = 2/36
P(A=4) = 3/36
P(A=5) = 4/36
P(A=6) = 5/36
P(A=7) = 6/36
P(A=8) = 5/36
P(A=9) = 4/36
P(A=10) = 3/36
P(A=11) = 2/36
P(A=12) = 1/36
The expected value of the sum of the numbers on the upper side of the dice is given by:
E(X) = Σk P(X=k)
where k is the sum of the numbers on the upper side of the dice.
Using the probabilities calculated above, we can find that:
E(X) = 2×1/36 + 3×2/36 + 4×3/36 + 5×4/36 + 6×5/36 + 7×6/36 + 8×5/36 + 9×4/36 + 10×3/36 + 11×2/36 + 12×1/36
Simplifying this expression, we get:
E(X) = (2+3+4+5+6+7+8+9+10+11+12)/2
E(X) = 7
Therefore, the expected value of the sum of the numbers on the upper side of the dice is 7.
Hence, option B is the correct answer.
If probability of drawing a spade from a well-shuffled pack of playing cards is ¼ then the probability that of the card drawn from a well-shuffled pack of playing cards is ‘not a spade’ is
- a)1
- b)½
- c)¼
- d)¾
Correct answer is option 'D'. Can you explain this answer?
If probability of drawing a spade from a well-shuffled pack of playing cards is ¼ then the probability that of the card drawn from a well-shuffled pack of playing cards is ‘not a spade’ is
a)
1
b)
½
c)
¼
d)
¾
|
|
Ipsita Rane answered |
Probability of Drawing a Spade
The probability of drawing a spade from a well-shuffled pack of playing cards is given as:
P(spade) =
where n(spade) is the number of spades in the deck and n(total) is the total number of cards in the deck.
Probability of Not Drawing a Spade
To find the probability of not drawing a spade, we need to subtract the probability of drawing a spade from 1 (since the probability of an event happening plus the probability of it not happening equals 1). Therefore, the probability of not drawing a spade is:
P(not spade) = 1 - P(spade)
Substituting the value of P(spade) in the above equation, we get:
P(not spade) = 1 -
Simplifying the expression, we get:
P(not spade) =
Therefore, the correct answer is option 'D'
The probability of drawing a spade from a well-shuffled pack of playing cards is given as:
P(spade) =
where n(spade) is the number of spades in the deck and n(total) is the total number of cards in the deck.
Probability of Not Drawing a Spade
To find the probability of not drawing a spade, we need to subtract the probability of drawing a spade from 1 (since the probability of an event happening plus the probability of it not happening equals 1). Therefore, the probability of not drawing a spade is:
P(not spade) = 1 - P(spade)
Substituting the value of P(spade) in the above equation, we get:
P(not spade) = 1 -
Simplifying the expression, we get:
P(not spade) =
Therefore, the correct answer is option 'D'
For a group of students, 30 %, 40% and 50% failed in Physics , Chemistry and at least one of the two subjects respectively. If an examinee is selected at random, what is the probability that he passed in Physics if it is known that he failed in Chemistry?- a)1/2
- b)1/3
- c)1/4
- d)1/6
Correct answer is option 'D'. Can you explain this answer?
For a group of students, 30 %, 40% and 50% failed in Physics , Chemistry and at least one of the two subjects respectively. If an examinee is selected at random, what is the probability that he passed in Physics if it is known that he failed in Chemistry?
a)
1/2
b)
1/3
c)
1/4
d)
1/6
|
|
Simran Pillai answered |
Let the total number of students = 100
Number of students failed in physics = 30% of 100 =30
Number of students failed in chemistry = 40% of 100 =40
Number of students failed at least one of the two subjects = 50% of 100 =50
We need to calculate
When a die is tossed, the sample space is- a)S =(1,2,3,4,5)
- b)S =(1,2,3,4)
- c)S =(1,2,3,4,5,6)
- d)none
Correct answer is option 'C'. Can you explain this answer?
When a die is tossed, the sample space is
a)
S =(1,2,3,4,5)
b)
S =(1,2,3,4)
c)
S =(1,2,3,4,5,6)
d)
none
|
|
Charvi Roy answered |
Explanation:
When a die is tossed, the possible outcomes are the numbers that can appear on the die. These numbers are from 1 to 6. Hence, the sample space is:
S = {1, 2, 3, 4, 5, 6}
Out of these six possible outcomes, only one can occur at a time. Therefore, the sample space is a set of mutually exclusive events.
Option 'A' is incorrect because it only lists five possible outcomes of the die. The sample space must include all possible outcomes.
Option 'B' is incorrect because it only lists four possible outcomes of the die. The sample space must include all possible outcomes.
Option 'D' is incorrect because none of the options listed include all possible outcomes of the die.
Therefore, the correct answer is option 'C' because it lists all possible outcomes of the die.
Conclusion:
The sample space of an experiment is the set of all possible outcomes of that experiment. In this case, the sample space of tossing a die is {1, 2, 3, 4, 5, 6}. It is important to identify the sample space correctly in order to calculate probabilities of various events.
When a die is tossed, the possible outcomes are the numbers that can appear on the die. These numbers are from 1 to 6. Hence, the sample space is:
S = {1, 2, 3, 4, 5, 6}
Out of these six possible outcomes, only one can occur at a time. Therefore, the sample space is a set of mutually exclusive events.
Option 'A' is incorrect because it only lists five possible outcomes of the die. The sample space must include all possible outcomes.
Option 'B' is incorrect because it only lists four possible outcomes of the die. The sample space must include all possible outcomes.
Option 'D' is incorrect because none of the options listed include all possible outcomes of the die.
Therefore, the correct answer is option 'C' because it lists all possible outcomes of the die.
Conclusion:
The sample space of an experiment is the set of all possible outcomes of that experiment. In this case, the sample space of tossing a die is {1, 2, 3, 4, 5, 6}. It is important to identify the sample space correctly in order to calculate probabilities of various events.
Sum of all probabilities is equal to- a)0
- b)½
- c)¾
- d)1
Correct answer is option 'D'. Can you explain this answer?
Sum of all probabilities is equal to
a)
0
b)
½
c)
¾
d)
1
|
|
Deepika Desai answered |
The sum of all probabilities is always equal to 1. This is a fundamental rule in probability theory and is known as the Law of Total Probability.
Explanation:
When we talk about probabilities, we are essentially talking about the likelihood or chance of an event occurring. Probability is always expressed as a number between 0 and 1, where 0 means that the event is impossible and 1 means that the event is certain.
For example, if we toss a fair coin, the probability of getting heads is 0.5 (or 50%) and the probability of getting tails is also 0.5 (or 50%). Together, these probabilities add up to 1.
The Law of Total Probability states that if we have a set of mutually exclusive and exhaustive events (i.e., events that cannot occur at the same time and that cover all possible outcomes), then the sum of the probabilities of these events is equal to 1.
For example, if we roll a fair six-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. Each of these outcomes is mutually exclusive (i.e., we can only get one of them at a time) and exhaustive (i.e., they cover all possible outcomes). Therefore, the sum of the probabilities of these outcomes is 1:
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
In this case, each outcome has an equal probability of 1/6, so we can rewrite the equation as:
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1
Which simplifies to:
6/6 = 1
This shows that the sum of all probabilities is always equal to 1, regardless of the number of events or their probabilities.
Explanation:
When we talk about probabilities, we are essentially talking about the likelihood or chance of an event occurring. Probability is always expressed as a number between 0 and 1, where 0 means that the event is impossible and 1 means that the event is certain.
For example, if we toss a fair coin, the probability of getting heads is 0.5 (or 50%) and the probability of getting tails is also 0.5 (or 50%). Together, these probabilities add up to 1.
The Law of Total Probability states that if we have a set of mutually exclusive and exhaustive events (i.e., events that cannot occur at the same time and that cover all possible outcomes), then the sum of the probabilities of these events is equal to 1.
For example, if we roll a fair six-sided die, there are six possible outcomes: 1, 2, 3, 4, 5, or 6. Each of these outcomes is mutually exclusive (i.e., we can only get one of them at a time) and exhaustive (i.e., they cover all possible outcomes). Therefore, the sum of the probabilities of these outcomes is 1:
P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
In this case, each outcome has an equal probability of 1/6, so we can rewrite the equation as:
1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1
Which simplifies to:
6/6 = 1
This shows that the sum of all probabilities is always equal to 1, regardless of the number of events or their probabilities.
If two events A and B are independent, then- a)They can be mutually exclusive
- b)They can not be mutually exclusive
- c)They can not be exhaustive
- d)Both (b) and (c)
Correct answer is option 'B'. Can you explain this answer?
If two events A and B are independent, then
a)
They can be mutually exclusive
b)
They can not be mutually exclusive
c)
They can not be exhaustive
d)
Both (b) and (c)
|
Dhruv Goyal answered |
B
As the events are independent means they cannot happen jointly ,so they are not mutually exclusive
As the events are independent means they cannot happen jointly ,so they are not mutually exclusive
An event that can be split into further events is known as- a)Complex event
- b)Mixed event
- c)Simple event
- d)Composite event
Correct answer is option 'D'. Can you explain this answer?
An event that can be split into further events is known as
a)
Complex event
b)
Mixed event
c)
Simple event
d)
Composite event
|
Arjun Singhania answered |
A composite event is an event made up of a grouping of elementary events, the possible outcomes of all three coins tossed together is a composite event. The event set is the possible outcome of an event. It can have the following values: (1) A single element: Queen of Hearts from a deck of cards: {Q }
If P(A/B) = P(A), then- a)A is independent of B
- b)B is independent of A
- c)B is dependent of A
- d)Both (a) and (b)
Correct answer is option 'D'. Can you explain this answer?
If P(A/B) = P(A), then
a)
A is independent of B
b)
B is independent of A
c)
B is dependent of A
d)
Both (a) and (b)
|
|
Mrinalini Iyer answered |
Explanation:
The given statement P(A/B) = P(A) can be rewritten as P(A ∩ B) = P(A)P(B).
If P(A/B) = P(A), then it means that the occurrence of event B does not affect the probability of event A. This can happen only if events A and B are independent of each other.
Therefore, the correct answer is option D, i.e. A is independent of B and B is independent of A.
In other words, if P(A/B) = P(A), then:
- Event A and event B are independent of each other.
- The occurrence of event B does not affect the probability of event A, and vice versa.
- The probability of A given B is equal to the probability of A, and vice versa.
The given statement P(A/B) = P(A) can be rewritten as P(A ∩ B) = P(A)P(B).
If P(A/B) = P(A), then it means that the occurrence of event B does not affect the probability of event A. This can happen only if events A and B are independent of each other.
Therefore, the correct answer is option D, i.e. A is independent of B and B is independent of A.
In other words, if P(A/B) = P(A), then:
- Event A and event B are independent of each other.
- The occurrence of event B does not affect the probability of event A, and vice versa.
- The probability of A given B is equal to the probability of A, and vice versa.
If variance of a random variable x is 23, then what is the variance of 2x+10?- a)56
- b)33
- c)46
- d)92
Correct answer is option 'D'. Can you explain this answer?
If variance of a random variable x is 23, then what is the variance of 2x+10?
a)
56
b)
33
c)
46
d)
92
|
|
Ishan Goyal answered |
Solution:
Given, Variance of a random variable x is 23.
We need to find the variance of 2x + 10.
Step 1: Find the mean of 2x + 10.
Mean of 2x + 10 = 2(mean of x) + 10
As we do not know the mean of x, we cannot find the exact mean of 2x + 10.
Step 2: Find the variance of 2x + 10.
Variance of 2x + 10 = Variance of 2x
= 4(Variance of x)
= 4(23)
= 92
Therefore, the variance of 2x + 10 is 92.
Answer: Option D (92)
Given, Variance of a random variable x is 23.
We need to find the variance of 2x + 10.
Step 1: Find the mean of 2x + 10.
Mean of 2x + 10 = 2(mean of x) + 10
As we do not know the mean of x, we cannot find the exact mean of 2x + 10.
Step 2: Find the variance of 2x + 10.
Variance of 2x + 10 = Variance of 2x
= 4(Variance of x)
= 4(23)
= 92
Therefore, the variance of 2x + 10 is 92.
Answer: Option D (92)
Which of the following set of function define a probability space on S = a1, a2, a3- a)P(a1)= 1/3, P(a2) = ½, P(a3)= ¼
- b)P(a1)= 1/3, P(a2)= 1/6,P(a3)= ½
- c)P(a1)= P(a2)= 2/3, P(a3)= ¼
- d)None
Correct answer is option 'B'. Can you explain this answer?
Which of the following set of function define a probability space on S = a1, a2, a3
a)
P(a1)= 1/3, P(a2) = ½, P(a3)= ¼
b)
P(a1)= 1/3, P(a2)= 1/6,P(a3)= ½
c)
P(a1)= P(a2)= 2/3, P(a3)= ¼
d)
None
|
|
Pallabi Khanna answered |
Probability Space on S = a1, a2, a3
A probability space is a mathematical construct that defines a set of possible outcomes in an experiment, along with their associated probabilities. In this question, we are given a set S = {a1, a2, a3} and asked to determine which set of functions defines a probability space on S.
Option A: P(a1)= 1/3, P(a2) = , P(a3)=
This option is incomplete as the probability of a2 is not given. A probability space must assign probabilities to all possible outcomes in the sample space.
Option B: P(a1)= 1/3, P(a2)= 1/6, P(a3)=
This option defines a probability space on S as it assigns probabilities to all three outcomes in the sample space. The probabilities are:
- P(a1) = 1/3
- P(a2) = 1/6
- P(a3) = 1 - P(a1) - P(a2) = 1/2
The sum of these probabilities is 1, which is a requirement for a valid probability space.
Option C: P(a1)= P(a2)= 2/3, P(a3)=
This option violates the requirement that the sum of all probabilities must be 1. The sum of P(a1) and P(a2) is 4/3, which is greater than 1.
Option D: None
Option D is not a valid option as it does not define any set of functions.
Conclusion
Option B is the correct answer as it defines a valid probability space on S. Option A is incomplete, option C violates the sum of probabilities requirement, and option D is not a valid option.
A probability space is a mathematical construct that defines a set of possible outcomes in an experiment, along with their associated probabilities. In this question, we are given a set S = {a1, a2, a3} and asked to determine which set of functions defines a probability space on S.
Option A: P(a1)= 1/3, P(a2) = , P(a3)=
This option is incomplete as the probability of a2 is not given. A probability space must assign probabilities to all possible outcomes in the sample space.
Option B: P(a1)= 1/3, P(a2)= 1/6, P(a3)=
This option defines a probability space on S as it assigns probabilities to all three outcomes in the sample space. The probabilities are:
- P(a1) = 1/3
- P(a2) = 1/6
- P(a3) = 1 - P(a1) - P(a2) = 1/2
The sum of these probabilities is 1, which is a requirement for a valid probability space.
Option C: P(a1)= P(a2)= 2/3, P(a3)=
This option violates the requirement that the sum of all probabilities must be 1. The sum of P(a1) and P(a2) is 4/3, which is greater than 1.
Option D: None
Option D is not a valid option as it does not define any set of functions.
Conclusion
Option B is the correct answer as it defines a valid probability space on S. Option A is incomplete, option C violates the sum of probabilities requirement, and option D is not a valid option.
Probability of throwing an odd no with an ordinary six faced die is- a)1/2
- b)1
- c)–1/2
- d)0
Correct answer is option 'A'. Can you explain this answer?
Probability of throwing an odd no with an ordinary six faced die is
a)
1/2
b)
1
c)
–1/2
d)
0
|
|
Sahil Malik answered |
Probability of Throwing an Odd Number with an Ordinary Six-Faced Die
To understand the probability of throwing an odd number with an ordinary six-faced die, we need to first understand what is meant by a six-faced die.
Definition of a Six-Faced Die
A six-faced die is a cube-shaped object with each face representing a number from 1 to 6. When the die is rolled, the outcome is random and can be any number from 1 to 6.
Understanding Probability
Probability is the likelihood or chance of an event occurring. It is expressed as a fraction or a percentage. For example, if we toss a coin, the probability of getting heads is 1/2 or 50%.
Determining the Probability of Throwing an Odd Number
To determine the probability of throwing an odd number with an ordinary six-faced die, we need to first identify the number of odd numbers that can be thrown. The odd numbers that can be thrown with a six-faced die are 1, 3, and 5.
Next, we need to determine the total number of possible outcomes. With a six-faced die, the total number of possible outcomes is 6 (1, 2, 3, 4, 5, and 6).
Finally, we can determine the probability of throwing an odd number by dividing the number of odd numbers by the total number of possible outcomes.
Probability of throwing an odd number = Number of odd numbers / Total number of possible outcomes
Probability of throwing an odd number = 3 / 6
Probability of throwing an odd number = 1/2
Therefore, the correct answer is option 'A', which states that the probability of throwing an odd number with an ordinary six-faced die is 1/2.
To understand the probability of throwing an odd number with an ordinary six-faced die, we need to first understand what is meant by a six-faced die.
Definition of a Six-Faced Die
A six-faced die is a cube-shaped object with each face representing a number from 1 to 6. When the die is rolled, the outcome is random and can be any number from 1 to 6.
Understanding Probability
Probability is the likelihood or chance of an event occurring. It is expressed as a fraction or a percentage. For example, if we toss a coin, the probability of getting heads is 1/2 or 50%.
Determining the Probability of Throwing an Odd Number
To determine the probability of throwing an odd number with an ordinary six-faced die, we need to first identify the number of odd numbers that can be thrown. The odd numbers that can be thrown with a six-faced die are 1, 3, and 5.
Next, we need to determine the total number of possible outcomes. With a six-faced die, the total number of possible outcomes is 6 (1, 2, 3, 4, 5, and 6).
Finally, we can determine the probability of throwing an odd number by dividing the number of odd numbers by the total number of possible outcomes.
Probability of throwing an odd number = Number of odd numbers / Total number of possible outcomes
Probability of throwing an odd number = 3 / 6
Probability of throwing an odd number = 1/2
Therefore, the correct answer is option 'A', which states that the probability of throwing an odd number with an ordinary six-faced die is 1/2.
Chapter doubts & questions for Chapter 15: Probability - Quantitative Aptitude for CA Foundation 2025 is part of CA Foundation exam preparation. The chapters have been prepared according to the CA Foundation exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for CA Foundation 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Chapter 15: Probability - Quantitative Aptitude for CA Foundation in English & Hindi are available as part of CA Foundation exam.
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