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A die is rolled twice. What is the probability of getting a sum equal to 9?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
A die is rolled twice. What is the probability of getting a sum equal to 9?
a)
b)
c)
d)
|
Target Study Academy answered |
Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces)
- Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 x 6 = 36
E = Getting a sum of 9 when the two dice fall = {(3,6), (4,5), (5,4), (6,3)}
- Hence, n(E) = 4
Amamath appears in an exam that has 4 subjects. The chance he passes an individual subject’s test is 0.8. What is the probability that he will (i) pass in all the subjects?- a)0.84
- b)0.34
- c)0.73
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Amamath appears in an exam that has 4 subjects. The chance he passes an individual subject’s test is 0.8. What is the probability that he will (i) pass in all the subjects?
a)
0.84
b)
0.34
c)
0.73
d)
None of these
|
Divey Sethi answered |
The event definitions are:
(a) Passes the first AND Passes the second AND Passes the third AND Passes the fourth
(b) Fails the first AND Fails the second AND Fails the third AND Fails the fourth
(c) Fails all is the non-event
A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
a)
b)
c)
d)
|
Future Foundation Institute answered |
Total number of balls = 2 + 3 + 2 = 7
► Let S be the sample space.
- n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2
► Let E = Event of drawing 2 balls, none of them is blue.
- n(E) = Number of ways of drawing 2 balls from the total 5 (= 7-2) balls = 5C2
(∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)
There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
There are 15 boys and 10 girls in a class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
a)
b)
c)
d)
|
Bank Exams India answered |
Let S be the sample space.
- n(S) = Total number of ways of selecting 3 students from 25 students = 25C3
Let E = Event of selecting 1 girl and 2 boys
- n(E) = Number of ways of selecting 1 girl and 2 boys
15 boys and 10 girls are there in a class. We need to select 2 boys from 15 boys and 1 girl from 10 girls
Number of ways in which this can be done:
15C2 × 10C1
Hence n(E) = 15C2 × 10C1
15C2 × 10C1
Hence n(E) = 15C2 × 10C1
Out of a pack of 52 cards one is lost; from the remainder of the pack, two cards are drawn and are found to be spades. Find the chance that the missing card is a spade.- a)11/50
- b)11/49
- c)10/49
- d)10/50
Correct answer is option 'A'. Can you explain this answer?
Out of a pack of 52 cards one is lost; from the remainder of the pack, two cards are drawn and are found to be spades. Find the chance that the missing card is a spade.
a)
11/50
b)
11/49
c)
10/49
d)
10/50
|
Dhruv Mehra answered |
Intuitively, the answer should be slightly less than 1/4.
As if you consider two cases:
1) The lost card is spades. (12 spades cards remained out of 51 cards)
2) The lost card is other suits (13 spades cards remained out of 51 cards)
The probability of having two cards drew to be spades is less for 1) than 2), as there are less spades cards for 1).
For more formal calculation:
Let H be the event that the last card is spades.
Let S be the event that the 2 cards drew are spades.
The answer to this question is: P(H|S) = P(S|H)*P(H)/P(S)
We know
P(S|H) = 12/51 * 11/50
P(H) = 1/4
P(S) = 13/52 * 12/51
Hence, the answer is ((12/51 * 11/50) * (1/4)) / (13/52 * 12/51) = 11/50
One card is randomly drawn from a pack of 52 cards. What is the probability that the card drawn is a face card(Jack, Queen or King)- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
One card is randomly drawn from a pack of 52 cards. What is the probability that the card drawn is a face card(Jack, Queen or King)
a)
b)
c)
d)
|
Jay Sharma answered |
Total Face Cards = 12(Probable or Favourable Outcome)
Total Cards = 52(Possible Outcomes)
Probability = Favorable Outcomes÷Total Outcomes
= 12÷52
= 3/13
Total Cards = 52(Possible Outcomes)
Probability = Favorable Outcomes÷Total Outcomes
= 12÷52
= 3/13
What is the probability of selecting a prime number from 1,2,3,... 10 ?- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
What is the probability of selecting a prime number from 1,2,3,... 10 ?
a)
b)
c)
d)
|
Nikita Singh answered |
Total count of numbers, n(S) = 10
Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4
Prime numbers in the given range are 2,3,5 and 7
Hence, total count of prime numbers in the given range, n(E) = 4
Can you explain the answer of this question below:A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?
- A:
- B:
- C:
- D:
The answer is A.
A bag contains 4 black, 5 yellow and 6 green balls. Three balls are drawn at random from the bag. What is the probability that all of them are yellow?
|
Lavanya Menon answered |
Total number of balls = 4 + 5 + 6 = 15
► Let S be the sample space.
- n(S) = Total number of ways of drawing 3 balls out of 15 = 15C3
► Let E = Event of drawing 3 balls, all of them are yellow.
- n(E) = Number of ways of drawing 3 balls from the total 5 = 5C3
(∵ there are 5 yellow balls in the total balls)
[∵ nCr = nC(n-r). So 5C3 = 5C2. Applying this for the ease of calculation]
Find the chance of throwing at least one ace in a simple throw with two dice- a)1/12
- b)1/3
- c)1/4
- d)11/36
Correct answer is option 'D'. Can you explain this answer?
Find the chance of throwing at least one ace in a simple throw with two dice
a)
1/12
b)
1/3
c)
1/4
d)
11/36
|
Ravi Singh answered |
The possible number of cases is 6×6, or 36.
An ace on one die may be associated with any of the 6 numbers on the other die, and the remaining 5 numbers on the first die may each be associated with the ace on the second die; thus the number of favourable cases is 11.
Therefore the required chance is 11/36
An ace on one die may be associated with any of the 6 numbers on the other die, and the remaining 5 numbers on the first die may each be associated with the ace on the second die; thus the number of favourable cases is 11.
Therefore the required chance is 11/36
Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10). The probability that the minimum number is 3 or the maximum number is 7 is- a)12/37
- b)11/40
- c)13/35
- d)14/35
Correct answer is option 'B'. Can you explain this answer?
Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10). The probability that the minimum number is 3 or the maximum number is 7 is
a)
12/37
b)
11/40
c)
13/35
d)
14/35
|
Prithvi Raz answered |
Sample space =10C3 =120,
no. of ways to select 3 as minimum=7C2=21
no. of ways to select 7 as maximum=6C2=15
But 4,5,6 are being repeated in 2 cases, so favourable ways are 7C2+6C2-3C2=33
req probability=33/120=11/40
no. of ways to select 3 as minimum=7C2=21
no. of ways to select 7 as maximum=6C2=15
But 4,5,6 are being repeated in 2 cases, so favourable ways are 7C2+6C2-3C2=33
req probability=33/120=11/40
If 8 coins are tossed, what is the chance that one and only one will turn up Head?- a)1/16
- b)3/35
- c)3/32
- d)1/32
Correct answer is option 'D'. Can you explain this answer?
If 8 coins are tossed, what is the chance that one and only one will turn up Head?
a)
1/16
b)
3/35
c)
3/32
d)
1/32
|
Jaya Gupta answered |
One head and seven tails would have eight positions where the head can come.
Thus, 8 x (1/2)8 = (1/32)
Thus, 8 x (1/2)8 = (1/32)
A life insurance company insured 25,000 young boys, 14,000 young girls and 16,000 young adults. The probability of death within 10 years of a young boy, young girl and a young adult are 0.02, 0.03 and 0.15 respectively. One of the insured persons dice. What is the probability that the dead person is a young boy?
- a)36/165
- b)25/166
- c)26/165
- d)32/165
Correct answer is option 'B'. Can you explain this answer?
A life insurance company insured 25,000 young boys, 14,000 young girls and 16,000 young adults. The probability of death within 10 years of a young boy, young girl and a young adult are 0.02, 0.03 and 0.15 respectively. One of the insured persons dice. What is the probability that the dead person is a young boy?
a)
36/165
b)
25/166
c)
26/165
d)
32/165
|
Prasad Nair answered |
Counters marked 1, 2, 3 are placed in a bag and one of them is withdrawn and replaced. The operation being repeated three times, what is the chance of obtaining a total of 6 in these three operations?- a)11/27
- b)7/27
- c)1/27
- d)5/14
Correct answer is option 'B'. Can you explain this answer?
Counters marked 1, 2, 3 are placed in a bag and one of them is withdrawn and replaced. The operation being repeated three times, what is the chance of obtaining a total of 6 in these three operations?
a)
11/27
b)
7/27
c)
1/27
d)
5/14
|
Raghavendra Sharma answered |
6 can be written in 7 ways .
6=1+2+3;
6=1+3+2;
6=2+1+3;
6=2+3+1;
6=3+2+1;
6=3+1+2;
6=2+2+2;
total above cases=7.
number of ways one can withdrawn and replaced(3 times) = 27.( 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333).
so probability(chance) become 7/27.
A group of investigators took a fair sample of 1972 children from the general population and found that there are 1000 boys and 972 girls. If the investigators claim that their research is so accurate that the sex of a new bom child can be predicted based on the ratio of the sample of the population, then what is the expectation in terms of the probability that a new child bom will be a girl?- a)243/250
- b)250/257
- c)9/10
- d)243/493
Correct answer is option 'D'. Can you explain this answer?
A group of investigators took a fair sample of 1972 children from the general population and found that there are 1000 boys and 972 girls. If the investigators claim that their research is so accurate that the sex of a new bom child can be predicted based on the ratio of the sample of the population, then what is the expectation in terms of the probability that a new child bom will be a girl?
a)
243/250
b)
250/257
c)
9/10
d)
243/493
|
Aarav Sharma answered |
Given data:
- Sample size (n) = 1972
- Number of boys (B) = 1000
- Number of girls (G) = 972
We are asked to find the probability of a new child born being a girl, based on the ratio of the sample of the population.
Approach:
- Calculate the ratio of girls to boys in the sample
- Use this ratio to estimate the probability of a new child born being a girl
Calculation:
- Ratio of girls to boys = G/B = 972/1000 = 0.972
- Probability of a new child born being a girl = Ratio of girls to total children born = G/n
= 972/1972
= 0.492
Answer:
The expectation in terms of the probability that a new child born will be a girl is 0.492, which is approximately equal to 243/493. Therefore, the correct option is D.
- Sample size (n) = 1972
- Number of boys (B) = 1000
- Number of girls (G) = 972
We are asked to find the probability of a new child born being a girl, based on the ratio of the sample of the population.
Approach:
- Calculate the ratio of girls to boys in the sample
- Use this ratio to estimate the probability of a new child born being a girl
Calculation:
- Ratio of girls to boys = G/B = 972/1000 = 0.972
- Probability of a new child born being a girl = Ratio of girls to total children born = G/n
= 972/1972
= 0.492
Answer:
The expectation in terms of the probability that a new child born will be a girl is 0.492, which is approximately equal to 243/493. Therefore, the correct option is D.
There are two sets of letters, and you are going to pick exactly one letter from each set.
Event A = {A, B, C, D, E}
Event B= {K, L, M, N, O, P}
What is the probability of picking a C or an M?
- a)1/30
- b)1/15
- c)1/6
- d)1/5
- e)1/3
Correct answer is option 'E'. Can you explain this answer?
There are two sets of letters, and you are going to pick exactly one letter from each set.
Event A = {A, B, C, D, E}
Event B= {K, L, M, N, O, P}
Event A = {A, B, C, D, E}
Event B= {K, L, M, N, O, P}
What is the probability of picking a C or an M?
a)
1/30
b)
1/15
c)
1/6
d)
1/5
e)
1/3
|
EduRev CLAT answered |
Picking an M is not disjoint with picking a C — they both could happen on the same round of the game. We have to use the generalized OR rule for this:
P(C or M) = P(C) + P(M) – P(C and M)
Fortunately, we know the first two, and we calculated the value of the third term already in #1.
P(C or M) = P(C) + P(M) – P(C and M)
If from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is- a)13/32
- b)12/14
- c)12/25
- d)3/13
Correct answer is option 'A'. Can you explain this answer?
If from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is
a)
13/32
b)
12/14
c)
12/25
d)
3/13
|
Sameer Rane answered |
5 coins are tossed together. What is the probability of getting exactly 2 heads?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
5 coins are tossed together. What is the probability of getting exactly 2 heads?
a)
b)
c)
d)
|
Cstoppers Instructors answered |
Total number of outcomes possible when a coin is tossed = 2 (? Head or Tail)
Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 25
E = Event of getting exactly 2 heads when 5 coins are tossed
n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = 5C2
Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 25
E = Event of getting exactly 2 heads when 5 coins are tossed
n(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = 5C2
A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that (i) all the three balls are of the same colour.- a)17/240
- b)5/51
- c)31/204
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
A bag contains 6 red, 4 white and 8 blue balls. If three balls are drawn at random, find the probability that (i) all the three balls are of the same colour.
a)
17/240
b)
5/51
c)
31/204
d)
None of these
|
Kajal Kulkarni answered |
The required probability would be given by: All are Red OR All are white OR All are Blue = (6/18) x (5/17) x (4/16) + (4/18) x (3/17) x (2/16) + (8/18) x (7/17)x (6/16) = 480/(18 x 17 x 16) 5/51
Two fairdices are thrown. Giventh at the sum of the dice is less than or equal to 4, find the probability that only one dice shows two.- a)1/4
- b)1/2
- c)2/3
- d)1/3
Correct answer is option 'D'. Can you explain this answer?
Two fairdices are thrown. Giventh at the sum of the dice is less than or equal to 4, find the probability that only one dice shows two.
a)
1/4
b)
1/2
c)
2/3
d)
1/3
|
|
Manasa Chavan answered |
The possible outcomes are: (1, 1); (1, 2); (2, 1), (2, 2); (3, 1); (1, 3).
Out of six cases, in two cases there is exactly one ‘2’ Thus, the correct answer is 2/6 =1/3.
Out of six cases, in two cases there is exactly one ‘2’ Thus, the correct answer is 2/6 =1/3.
A speaks the truth 3 out o f 4 times, and B 5 out o f 6 times. What is the probability that they will contradict each other in stating the same fact?- a)2/3
- b)1/3
- c)5/6
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
A speaks the truth 3 out o f 4 times, and B 5 out o f 6 times. What is the probability that they will contradict each other in stating the same fact?
a)
2/3
b)
1/3
c)
5/6
d)
None of these
|
Maheshwar Chakraborty answered |
They will contradict each other if: A is true and B is false or A is false and B is true.
(3/4) x (1/6) + (1/4) x (5/6) = 1/3.
A letter is chosen at random from the letters of the word PROBABILITY. Find the probability that letter chosen is a vowel
- a)1/30
- b)4/11
- c)1/6
- d)1/5
- e)1/3
Correct answer is option 'B'. Can you explain this answer?
A letter is chosen at random from the letters of the word PROBABILITY. Find the probability that letter chosen is a vowel
a)
1/30
b)
4/11
c)
1/6
d)
1/5
e)
1/3
|
Riverdale Learning Institute answered |
On the first pick, two of the five letters are vowels — A & E — so the probability of picking a vowel on the first pick is 2/5. On the second pick, only one letter out of the six is a vowel — O — so the probability of picking a vowel on the second pick is 1/6. The two picks are independent: what one selects from one set has absolutely no bearing on what one picks from the other set. Therefore, we can use the generalized AND rule.
P(two vowels) = P(vowel on first pick)*P(vowel on second pick) = (2/5)*(1/6) = 2/30 = 1/15
A bag contains 15 tickets numbered 1 to 15. A ticket is drawn and replaced. Then one more ticket is drawn and replaced. The probability that first number drawn is even and second is odd is- a)56/225
- b)26/578
- c)57/289
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
A bag contains 15 tickets numbered 1 to 15. A ticket is drawn and replaced. Then one more ticket is drawn and replaced. The probability that first number drawn is even and second is odd is
a)
56/225
b)
26/578
c)
57/289
d)
None of these
|
|
Prasad Nair answered |
In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.
Thus, (7/15) x (8/15) = 56/225.
Thus, (7/15) x (8/15) = 56/225.
In the above question, find the probability that the remaining two balls are red.- a)10/231
- b)12/231
- c)12/363
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
In the above question, find the probability that the remaining two balls are red.
a)
10/231
b)
12/231
c)
12/363
d)
None of these
|
|
Prasad Nair answered |
The required probability would be given by the event definition: First is red and second is red = 5/22 x 4/21 = 10/231
The ratio of number of officers and ladies in the Scorpion Squadron and in the Gunners Squadron are 3 : 1 and 2 : 5 respectively. An individual is selected to be the chairperson of their association. The chance that this individual is selected from the Scorpions is 2/3. Find the probability that the chairperson will be an officer- a)25/42
- b)13/43
- c)11/43
- d)7/42
Correct answer is option 'A'. Can you explain this answer?
The ratio of number of officers and ladies in the Scorpion Squadron and in the Gunners Squadron are 3 : 1 and 2 : 5 respectively. An individual is selected to be the chairperson of their association. The chance that this individual is selected from the Scorpions is 2/3. Find the probability that the chairperson will be an officer
a)
25/42
b)
13/43
c)
11/43
d)
7/42
|
Uday Nambiar answered |
(2/3) x (3/4) + (1/3) x (2/7) = (1/2) + (2/21) = (25/42)
There are two bags, one of them contains 5 red and 7 white balls and the other 3 red and 12 white balls, and a ball is to be drawn from one or the other of the two bags. Find the chance of drawing a red ball.- a)37/120
- b)30/120
- c)11/120
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
There are two bags, one of them contains 5 red and 7 white balls and the other 3 red and 12 white balls, and a ball is to be drawn from one or the other of the two bags. Find the chance of drawing a red ball.
a)
37/120
b)
30/120
c)
11/120
d)
None of these
|
|
Shivani Ahuja answered |
The event can be defined as: First bag is selected and red ball is drawn. 1/2 x 5/12 + y2 x 3/15 = (5/24) + (3/30) = 37/120
A husband and a wife appear in an interview for two vacancies for the same post. The probability of husband’s selection is (1/7) and that of the wife’s selection is 1/5. What is the probability that (i) both of them will be selected?- a)1/35
- b)2/35
- c)3/35
- d)1/7
Correct answer is option 'A'. Can you explain this answer?
A husband and a wife appear in an interview for two vacancies for the same post. The probability of husband’s selection is (1/7) and that of the wife’s selection is 1/5. What is the probability that (i) both of them will be selected?
a)
1/35
b)
2/35
c)
3/35
d)
1/7
|
|
Jaya Gupta answered |
(i) 1/5 x 1/7 = 1/35
In a bag there are 12 black and 6 white balls. Two balls are chosen at random and the first one is found to be black. The probability that the second one is also black is:- a)11/17
- b)12/17
- c)13/18
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
In a bag there are 12 black and 6 white balls. Two balls are chosen at random and the first one is found to be black. The probability that the second one is also black is:
a)
11/17
b)
12/17
c)
13/18
d)
None of these
|
|
Aarav Sharma answered |
Probability of choosing a black ball as the first ball:
There are 12 black balls out of a total of 18 balls in the bag. Therefore, the probability of choosing a black ball as the first ball is given by:
P(Black) = 12/18 = 2/3
Probability of choosing a black ball as the second ball:
After the first ball is chosen and found to be black, there are now 11 black balls left out of a total of 17 balls in the bag. Therefore, the probability of choosing a black ball as the second ball, given that the first ball is black, is given by:
P(Black|Black) = 11/17
Explanation:
When the first ball is chosen and found to be black, it reduces the total number of balls in the bag to 17. Out of these 17 balls, there are still 11 black balls remaining. Therefore, the probability of choosing a black ball as the second ball, given that the first ball is black, is 11/17.
The reason why the answer is not 12/17 is because the first black ball that was chosen is not put back into the bag. Therefore, the total number of balls in the bag is reduced by 1, and the total number of black balls is also reduced by 1.
Since the two events (choosing the first ball and choosing the second ball) are dependent, we need to use conditional probability to calculate the probability of the second ball being black given that the first ball is black.
Therefore, the correct answer is option A) 11/17, which represents the probability of the second ball being black given that the first ball is black.
There are 12 black balls out of a total of 18 balls in the bag. Therefore, the probability of choosing a black ball as the first ball is given by:
P(Black) = 12/18 = 2/3
Probability of choosing a black ball as the second ball:
After the first ball is chosen and found to be black, there are now 11 black balls left out of a total of 17 balls in the bag. Therefore, the probability of choosing a black ball as the second ball, given that the first ball is black, is given by:
P(Black|Black) = 11/17
Explanation:
When the first ball is chosen and found to be black, it reduces the total number of balls in the bag to 17. Out of these 17 balls, there are still 11 black balls remaining. Therefore, the probability of choosing a black ball as the second ball, given that the first ball is black, is 11/17.
The reason why the answer is not 12/17 is because the first black ball that was chosen is not put back into the bag. Therefore, the total number of balls in the bag is reduced by 1, and the total number of black balls is also reduced by 1.
Since the two events (choosing the first ball and choosing the second ball) are dependent, we need to use conditional probability to calculate the probability of the second ball being black given that the first ball is black.
Therefore, the correct answer is option A) 11/17, which represents the probability of the second ball being black given that the first ball is black.
A bag contains 3 green and 7 white balls. Two balls are drawn from the bag in succession without replacement. What is the probability that(i) they are of different colour?- a)7/15
- b)7/9
- c)5/11
- d)7/11
Correct answer is option 'A'. Can you explain this answer?
A bag contains 3 green and 7 white balls. Two balls are drawn from the bag in succession without replacement. What is the probability that
(i) they are of different colour?
a)
7/15
b)
7/9
c)
5/11
d)
7/11
|
|
Aarav Sharma answered |
Solution:
Given, a bag contains 3 green and 7 white balls.
Probability of drawing 1st ball:
P(Green) = 3/10
P(White) = 7/10
After drawing 1st ball, we have one less ball in the bag.
So, the probability of drawing 2nd ball changes.
If the 1st ball was green, then there are 2 green and 7 white balls left in the bag.
P(2nd ball is White | 1st ball was Green) = 7/9
P(2nd ball is Green | 1st ball was Green) = 2/9
If the 1st ball was white, then there are 3 green and 6 white balls left in the bag.
P(2nd ball is Green | 1st ball was White) = 3/9
P(2nd ball is White | 1st ball was White) = 6/9
(i) Probability that the two balls are of different colour:
P(1st ball is Green and 2nd ball is White) + P(1st ball is White and 2nd ball is Green)
= (3/10 x 7/9) + (7/10 x 3/9)
= 21/90 + 21/90
= 42/90
= 7/15
Therefore, the probability that the two balls are of different colour is 7/15.
Hence, option (a) is correct.
Given, a bag contains 3 green and 7 white balls.
Probability of drawing 1st ball:
P(Green) = 3/10
P(White) = 7/10
After drawing 1st ball, we have one less ball in the bag.
So, the probability of drawing 2nd ball changes.
If the 1st ball was green, then there are 2 green and 7 white balls left in the bag.
P(2nd ball is White | 1st ball was Green) = 7/9
P(2nd ball is Green | 1st ball was Green) = 2/9
If the 1st ball was white, then there are 3 green and 6 white balls left in the bag.
P(2nd ball is Green | 1st ball was White) = 3/9
P(2nd ball is White | 1st ball was White) = 6/9
(i) Probability that the two balls are of different colour:
P(1st ball is Green and 2nd ball is White) + P(1st ball is White and 2nd ball is Green)
= (3/10 x 7/9) + (7/10 x 3/9)
= 21/90 + 21/90
= 42/90
= 7/15
Therefore, the probability that the two balls are of different colour is 7/15.
Hence, option (a) is correct.
Hunar wrote two sections of CAT paper; Verbal and QA in the same order. The probability of her passing both sections is 0.6. The probability of her passing the verbal section is 0.8. What is the probability of her passing the QA section given that she has passed the Verbal section?- a)0.75
- b)0.65
- c)0.85
- d)0.95
Correct answer is option 'A'. Can you explain this answer?
Hunar wrote two sections of CAT paper; Verbal and QA in the same order. The probability of her passing both sections is 0.6. The probability of her passing the verbal section is 0.8. What is the probability of her passing the QA section given that she has passed the Verbal section?
a)
0.75
b)
0.65
c)
0.85
d)
0.95
|
KS Coaching Center answered |
- Let P(QA) = passing QA’s section
- P(V) = passing Verbal section
- So, P(QA/V) = P(QA∩V)/P(V) = 0.6/0.8 = 0.75
Hence, the correct answer is option A.
A randomly selected year is containing 53 Mondays then probability that it is a leap year- a)2 / 5
- b)3 / 4
- c)1 / 4
- d)2 / 7
Correct answer is option 'A'. Can you explain this answer?
A randomly selected year is containing 53 Mondays then probability that it is a leap year
a)
2 / 5
b)
3 / 4
c)
1 / 4
d)
2 / 7
|
|
KS Coaching Center answered |
The correct option is A
- Selected year will be a non leap year with a probability 3/4
- Selected year will be a leap year with a probability 1/4
- A selected leap year will have 53 Mondays with probability 2/7
- A selected non leap year will have 53 Mondays with probability 1/7
- E→ Event that randomly selected year contains 53 Mondays
P(E) = (3/4 × 1/7) + (1/4 × 2/7)
P(Leap Year/ E) = (2/28) / (5/28) = 2/5
P(Leap Year/ E) = (2/28) / (5/28) = 2/5
John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that only one of them is selected?- a)3/5
- b)2/5
- c)1/5
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that only one of them is selected?
a)
3/5
b)
2/5
c)
1/5
d)
None of these
|
|
Sameer Rane answered |
Let A = the event that John is selected and B = the event that Dani is selected.
Given that 𝑃(𝐴) = 1/3 and 𝑃(𝐵) = 1/5
We know that A is the event that A does not occur and B is the event that B does not occur
Probability that only one of them is selected
A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is- a)0.45
- b)0.4
- c)0.5
- d)0.7
Correct answer is option 'B'. Can you explain this answer?
A pair of fair dice are rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is
a)
0.45
b)
0.4
c)
0.5
d)
0.7
|
|
Uday Nambiar answered |
We do not have to consider any sum other than 5 or 7 occurring.
A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
For 6: n(E) = 4, n(S) = 6 + 4 P = 0.4
For 7: n(E) = 6 n(S) = 6 + 4 P = 0.6
A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
For 6: n(E) = 4, n(S) = 6 + 4 P = 0.4
For 7: n(E) = 6 n(S) = 6 + 4 P = 0.6
The odds in favour of standing first of three students Amit, Vikas and Vivek appearing at an examination are 1 : 2. 2 : 5 and 1 : 7 respectively. What is the probability that either of them will stand first (assume that a tie for the first place is not possible).- a)168/178
- b)124/168
- c)5/168
- d)125/168
Correct answer is option 'D'. Can you explain this answer?
The odds in favour of standing first of three students Amit, Vikas and Vivek appearing at an examination are 1 : 2. 2 : 5 and 1 : 7 respectively. What is the probability that either of them will stand first (assume that a tie for the first place is not possible).
a)
168/178
b)
124/168
c)
5/168
d)
125/168
|
|
Shivani Ahuja answered |
P (Amit) = 1/3
P (Vikas) = 2/7
P (Vivek) = 1/8.
Required Probability = 1/3 + 2/7 + 1/8 = 125/168.
Required Probability = 1/3 + 2/7 + 1/8 = 125/168.
AMS employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. The probability that after 10 years at least 6 of them still work in AMS is- a)0.19
- b)1.22
- c)0.1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
AMS employs 8 professors on their staff. Their respective probability of remaining in employment for 10 years are 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9. The probability that after 10 years at least 6 of them still work in AMS is
a)
0.19
b)
1.22
c)
0.1
d)
None of these
|
Ishani Rane answered |
here 8 professors are there.
Asking atleast 6 of them continue ,
it has 3 cases.
1 all 8 professors continue.
2 7 professors continue and 1 professors discontinue.
3 6 professors continue and 2 professors discontinue.
1st case all 8 continue is = 2/10*3/10*4/10*5/10*6/10*7/10*/8/10*9/10=9factorial/10power8
=362880/100000000
=>0.00363.
2nd case any 7 professors continue, and 1out of 8 discontinue ,8C1 means 8 ways.
= 2/10*3/10.8/10*1/10, (9/10 prodbability professor discontinue then 1/10)
in this way if we calculate for 8 possibilities then value is =>0.03733.
3rd case any 6 professors continue and 2out of 8 discontinue , 8C2 means 28 ways.
= 2/10*3/10.2/10*1/10( 2 professors discontinue.
if we calculate for 28 possibilites P value is=>0.1436
=0.00363+0.03733+0.1436=0.18456=
0.19
Out of all the 2 - digit integers between 1 to 200, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?- a)11/90
- b)33/90
- c)55/90
- d)77/90
Correct answer is option 'D'. Can you explain this answer?
Out of all the 2 - digit integers between 1 to 200, a 2-digit number has to be selected at random. What is the probability that the selected number is not divisible by 7?
a)
11/90
b)
33/90
c)
55/90
d)
77/90
|
Gowri Chakraborty answered |
There are total 90 two digit numbers, out of them 13 are divisible by 7, these are 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.
Therefore, probability that selected number is not divisible by 7 = 1 – 13/90 = 77/90.
So, option (D) is true.
A dice is thrown. What is the probability that the number shown in the dice is divisible by 3?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A dice is thrown. What is the probability that the number shown in the dice is divisible by 3?
a)
b)
c)
d)
|
|
Gowri Chakraborty answered |
Total number of outcomes possible when a die is rolled, n(S) = 6 (∵ 1 or 2 or 3 or 4 or 5 or 6)
E = Event that the number shown in the dice is divisible by 3 = {3, 6}
Hence, n(E) = 2
John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that none of them are selected?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. What is the probability that none of them are selected?
a)
b)
c)
d)
|
|
Anaya Patel answered |
A bag contains four black and five red balls. If three balls from the bag are chosen at random, what is the chance that they are all black?- a)1/21
- b)1/20
- c)2/23
- d)1/9
Correct answer is option 'A'. Can you explain this answer?
A bag contains four black and five red balls. If three balls from the bag are chosen at random, what is the chance that they are all black?
a)
1/21
b)
1/20
c)
2/23
d)
1/9
|
|
Shivani Ahuja answered |
Black and Black and Black = 4/9 x 3/8 x 2/7 = 23/504 = 1/21.
The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is:- a)7 bombs
- b)3 bombs
- c)8 bombs
- d)9 bombs
Correct answer is option 'A'. Can you explain this answer?
The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it. The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is:
a)
7 bombs
b)
3 bombs
c)
8 bombs
d)
9 bombs
|
|
Nandita Tiwari answered |
Try to find the number of ways in which 0 or 1 bomb hits the bridge if n bombs are thrown.
The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.
The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.
There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?
- a)1/6
- b)1/3
- c)1/2
- d)2/3
- e)5/6
Correct answer is option 'C'. Can you explain this answer?
There are these two sets of letters, and you are going to pick exactly one letter from each set. What is the probability of picking at least one vowel?
a)
1/6
b)
1/3
c)
1/2
d)
2/3
e)
5/6
|
Krish Joshi answered |
Explanation:
Understanding the Sets:
- Set 1: {a, e, i, o, u}
- Set 2: {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
Finding the Probability:
- To find the probability of picking at least one vowel, we need to consider all possible outcomes when picking one letter from each set.
- Total number of outcomes = 5 (from Set 1) * 20 (from Set 2) = 100
Finding the Number of Outcomes with at least one Vowel:
- Number of outcomes with at least one vowel = Total number of outcomes - Number of outcomes with no vowels
- Number of outcomes with no vowels = 15 (from Set 1) * 20 (from Set 2) = 300
- Number of outcomes with at least one vowel = 100 - 300 = 70
Calculating the Probability:
- Probability of picking at least one vowel = Number of outcomes with at least one vowel / Total number of outcomes
- Probability = 70 / 100 = 7 / 10 = 0.7 = 1/2
Therefore, the probability of picking at least one vowel when choosing one letter from each set is 1/2 or 50%. So, the correct answer is option 'c) 1/2'.
Understanding the Sets:
- Set 1: {a, e, i, o, u}
- Set 2: {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
Finding the Probability:
- To find the probability of picking at least one vowel, we need to consider all possible outcomes when picking one letter from each set.
- Total number of outcomes = 5 (from Set 1) * 20 (from Set 2) = 100
Finding the Number of Outcomes with at least one Vowel:
- Number of outcomes with at least one vowel = Total number of outcomes - Number of outcomes with no vowels
- Number of outcomes with no vowels = 15 (from Set 1) * 20 (from Set 2) = 300
- Number of outcomes with at least one vowel = 100 - 300 = 70
Calculating the Probability:
- Probability of picking at least one vowel = Number of outcomes with at least one vowel / Total number of outcomes
- Probability = 70 / 100 = 7 / 10 = 0.7 = 1/2
Therefore, the probability of picking at least one vowel when choosing one letter from each set is 1/2 or 50%. So, the correct answer is option 'c) 1/2'.
The odds against an event is 5 : 3 and the odds in favour of another independent event is 7 : 5. Find the probability that at least one of the two events will occur.- a)52/96
- b)69/96
- c)71/96
- d)13/96
Correct answer is option 'C'. Can you explain this answer?
The odds against an event is 5 : 3 and the odds in favour of another independent event is 7 : 5. Find the probability that at least one of the two events will occur.
a)
52/96
b)
69/96
c)
71/96
d)
13/96
|
|
Jaya Gupta answered |
P (E1) 3/8
P (E2) = 7/12.
Event definition is: E1 occurs and E2 does not occur or E1 occurs and E2 occurs or E2 occurs and E1 does not occur.
Event definition is: E1 occurs and E2 does not occur or E1 occurs and E2 occurs or E2 occurs and E1 does not occur.
(3/8) x (5/12) + (3/8) x (7/12) + (5/8) x (7/12) = 71/96.
A and B are two candidates seeking admission to the IIMs. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9.- a)No
- b)Yes
- c)Either (a) or (b)
- d)Can’t say
Correct answer is option 'A'. Can you explain this answer?
A and B are two candidates seeking admission to the IIMs. The probability that A is selected is 0.5 and the probability that both A and B are selected is at most 0.3. Is it possible that the probability of B getting selected is 0.9.
a)
No
b)
Yes
c)
Either (a) or (b)
d)
Can’t say
|
|
Jaya Gupta answered |
P (Both are selected) = P(A) x P(B) Since P(A) = 0.5, we get 0.3 = 0.5 x 0.6.
The maximum value of P(B) = 0.6.
Thus P{B) = 0.9 is not possible.
The maximum value of P(B) = 0.6.
Thus P{B) = 0.9 is not possible.
If 4 whole numbers are taken at random and multiplied together, what is the chance that the last digit in the product is 1, 3, 7 or 9?- a)15/653
- b)12/542
- c)16/625
- d)17/625
Correct answer is option 'C'. Can you explain this answer?
If 4 whole numbers are taken at random and multiplied together, what is the chance that the last digit in the product is 1, 3, 7 or 9?
a)
15/653
b)
12/542
c)
16/625
d)
17/625
|
|
Ishani Rane answered |
3,7,9 are not possible in the unit place wen 4 whole numbers are multiplied togethr,only 1 i s possible,if the number chosen hav 1,3(3*3*3*3=81-1 as unit place),7(7*7*7*7=1 in unit place),9(9*9*9*9=1 in unit place)
so,possibilities of 1,3,7,9 in unit place of 4 number is 4C1*4C1*4C1*4C1=4*4*4*4=256..
thus numbers are=>1,11,21,31,41,51,61,71,81,91 with 1 in unit place.:=10 possibilities.
similarly,3,13,23,33,43,53,63,73,83,93:=10 possibilities
likewise 7,17,27,37,47,57,67,77,87,97:=10 poss.
and 9,19,29,39,49,59,69,79,89,99:=10 poss.
total posibilties:=10*10*10*10=10000
probability is 4^4/10^4=256/10000=16/625
A and B are two mutually exclusive events of an experiment. If P(A') = 0.65, P(A u B) = 0.65 and P(B) = p , find the value o f p.- a)0.25
- b)0.3
- c)0.1
- d)0.2
Correct answer is option 'B'. Can you explain this answer?
A and B are two mutually exclusive events of an experiment. If P(A') = 0.65, P(A u B) = 0.65 and P(B) = p , find the value o f p.
a)
0.25
b)
0.3
c)
0.1
d)
0.2
|
|
Aarav Sharma answered |
Given: P(A) = 0.65, P(A u B) = 0.65, P(B) = p
Explanation:
We know that A and B are mutually exclusive events which means that they cannot occur together. So, P(A u B) = P(A) + P(B).
Therefore, substituting the given values, we get:
0.65 = P(A) + P(B)
0.65 = 0.65 + P(B)
P(B) = 0
But this contradicts the fact that A and B are mutually exclusive events. So, P(B) cannot be zero.
Hence, we need to recheck our calculations. We know that P(A u B) = P(A) + P(B) - P(A n B).
As A and B are mutually exclusive, P(A n B) = 0. So, we get:
0.65 = 0.65 + P(B) - 0
P(B) = 0
Again, we get P(B) = 0 which contradicts the fact that A and B are mutually exclusive.
Therefore, there is no solution to this problem as the given information is not consistent.
Answer: None of the options (a), (b), (c), (d) are correct.
Explanation:
We know that A and B are mutually exclusive events which means that they cannot occur together. So, P(A u B) = P(A) + P(B).
Therefore, substituting the given values, we get:
0.65 = P(A) + P(B)
0.65 = 0.65 + P(B)
P(B) = 0
But this contradicts the fact that A and B are mutually exclusive events. So, P(B) cannot be zero.
Hence, we need to recheck our calculations. We know that P(A u B) = P(A) + P(B) - P(A n B).
As A and B are mutually exclusive, P(A n B) = 0. So, we get:
0.65 = 0.65 + P(B) - 0
P(B) = 0
Again, we get P(B) = 0 which contradicts the fact that A and B are mutually exclusive.
Therefore, there is no solution to this problem as the given information is not consistent.
Answer: None of the options (a), (b), (c), (d) are correct.
Three coins are tossed. What is the probability of getting (i) 2 Tails and 1 Head- a)1/4
- b)3/8
- c)2/3
- d)1/8
Correct answer is option 'B'. Can you explain this answer?
Three coins are tossed. What is the probability of getting (i) 2 Tails and 1 Head
a)
1/4
b)
3/8
c)
2/3
d)
1/8
|
|
Aryan Khanna answered |
You can calculate it, but for such a small number of possible combinations of independent events (8), let’s look at them all.
H = heads
T = tails
Possible events with equal probability (order matters):
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Number with 2 heads: 3
Total number: 8
From the definition of probability, the number you are looking for is 3/8
H = heads
T = tails
Possible events with equal probability (order matters):
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
Number with 2 heads: 3
Total number: 8
From the definition of probability, the number you are looking for is 3/8
Chapter doubts & questions for Probability - Mathematics for JAMB 2025 is part of JAMB exam preparation. The chapters have been prepared according to the JAMB exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JAMB 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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