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All questions of Unit 4: Probability, Random Variables, and Probability Distributions for Grade 9 Exam
One card is drawn from a pack of cards, each of the 52 cards being equally likely to be drawn. The probability that the card drawn is red and a queen is:- a)1/26
- b)1/2
- c)2
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
One card is drawn from a pack of cards, each of the 52 cards being equally likely to be drawn. The probability that the card drawn is red and a queen is:
a)
1/26
b)
1/2
c)
2
d)
1/4
|
Gaurav Kumar answered |
The cards contains 4 Queen from which 2 are black and 2 are red
we need to find the probability that the card drawn is red and a queen is: 2/52
= 1/26
we need to find the probability that the card drawn is red and a queen is: 2/52
= 1/26
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Three coins are tossed. If at least two coins show head, the probability of getting one tail is:- a)3/4
- b)1/3
- c)1
- d)2/3
Correct answer is option 'A'. Can you explain this answer?
Three coins are tossed. If at least two coins show head, the probability of getting one tail is:
a)
3/4
b)
1/3
c)
1
d)
2/3
|
Amhe Nadi answered |
Sample space = HHH, HHT, HTH, THH, TTH, THT, HTT, TTTGetting atleast two heads(F) = HHH, HHT, HTH, THHProbability of getting atleast two heads(F) = 4/8 =1/2Getting one tail (E) = HHT, HTH, THHE intersection F = HHT, HTH, THHP(E intersection F) = 3/8Required probability = P(E intersection F) / P(F) = 3/8 ÷ 1/2 =3/4
The number of adults living in homes on a randomly selected city block is described by the following probability distribution.
What is the probability that 4 or more adults reside at a randomly selected home?- a).50
- b).25
- c).10
- d).15
Correct answer is option 'C'. Can you explain this answer?
The number of adults living in homes on a randomly selected city block is described by the following probability distribution.
What is the probability that 4 or more adults reside at a randomly selected home?
a)
.50
b)
.25
c)
.10
d)
.15
|
Praveen Kumar answered |
The sum of all the probabilities is equal to 1.
Therefore, the probability that four or more adults reside in a home = 1 - (0.25 + 0.50 + 0.15)
= 0.10.
Therefore, the probability that four or more adults reside in a home = 1 - (0.25 + 0.50 + 0.15)
= 0.10.
A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. The probability of occurrence of A is.- a)1/2
- b)1/6
- c)5/6
- d)1/3
Correct answer is option 'A'. Can you explain this answer?
A and B are two independent events. The probability that both A and B occur is 1/6 and the probability that neither of them occurs is 1/3. The probability of occurrence of A is.
a)
1/2
b)
1/6
c)
5/6
d)
1/3
|
Hansa Sharma answered |
The probability that both occur simultaneously is 1/6 and the probability that neither occurs is 1/3
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
On Solving for x and y,
we get x=1/3 or x=1/2 which is the probability of occurrence of A.
Let P(A)=x, P(B)=y
Then P(A)×P(B) = ⅙ becomes xy = 1/6
And [1−P(A)][1−P(B)]=1/3 becomes (1−x)(1−y)=1/3
On Solving for x and y,
we get x=1/3 or x=1/2 which is the probability of occurrence of A.
A bag contains 25 tickets numbered from 1 to 25. Two tickets are drawn one after another without replacement. The probability that both tickets will show even numbers is:- a)11/50
- b)11/24
- c)13/25
- d)3/8
Correct answer is 'A'. Can you explain this answer?
A bag contains 25 tickets numbered from 1 to 25. Two tickets are drawn one after another without replacement. The probability that both tickets will show even numbers is:
a)
11/50
b)
11/24
c)
13/25
d)
3/8
|
Tejas Verma answered |
There are 12 even numbers between 1 to 25.Consider the given events.A = An even number ticket in the first drawB = An even number ticket in the second drawNow ,
A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?- a)
- b)
- c) zero
- d)1
Correct answer is option 'A'. Can you explain this answer?
A box contains 15 red marbles, 15 blue marbles and 30 green marbles. 5 marbles are drawn from the box, what is the probability that atleast one will be green?
a)
b)
c)
zero
d)
1
|
Shashwat Singh answered |
It's like doing the question orally... (don't get into maths... as I was also stuck in the middle) but by common sense (which is very uncommon) it can be said as 1-probablity of not getting any green in 5 balls... in this case we can see the answer aproaches to 1 which can be seen in option a) .... hope it helps... but with maths a can see a nightmare coming...
Probability of getting a number between 1 and 100, which is divisible by 1 and itself only, is- a)25/99
- b)1/4
- c)25/98
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Probability of getting a number between 1 and 100, which is divisible by 1 and itself only, is
a)
25/99
b)
1/4
c)
25/98
d)
none of these
|
Nandini Iyer answered |
The prime numbers between 1 and 100 are :2, 3, 5, 7, 11,
The probability that a man will live for 10 more years is 1/4 and that his wife will live 10 more years is 1/3. The probability that neither will be alive in 10 years is- a)5/12
- b)1/2
- c)11/12
- d)7/12
Correct answer is option 'B'. Can you explain this answer?
The probability that a man will live for 10 more years is 1/4 and that his wife will live 10 more years is 1/3. The probability that neither will be alive in 10 years is
a)
5/12
b)
1/2
c)
11/12
d)
7/12
|
Mayank Dasgupta answered |
Given probabilities:
- Probability that the man will live for 10 more years = 1/4
- Probability that his wife will live for 10 more years = 1/3
To find: Probability that neither will be alive in 10 years
Solution:
Let A be the event that the man will be alive in 10 years, and B be the event that his wife will be alive in 10 years. Then, the probability that neither will be alive in 10 years is the probability of the complement of the union of A and B, 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 the man and his wife will be alive in 10 years.
Substituting the given probabilities, we get:
P(A ∪ B) = 1/4 + 1/3 - P(A ∩ B)
Simplifying, we get:
P(A ∩ B) = 7/12 - P((A ∪ B)')
Now, we know that the probability of the man being alive in 10 years is 1/4, which means the probability of him not being alive in 10 years is 3/4. Similarly, the probability of his wife not being alive in 10 years is 2/3.
Using the product rule of probability, we can find the probability of both of them not being alive in 10 years:
P((A ∪ B)') = P(A' ∩ B') = P(A') × P(B') = (3/4) × (2/3) = 1/2
Substituting this value in the above equation, we get:
P(A ∩ B) = 7/12 - 1/2 = 1/12
Therefore, the probability that neither the man nor his wife will be alive in 10 years is:
P((A ∪ B)') = 1 - P(A ∪ B) = 1 - (1/4 + 1/3 - 1/12) = 1/2
Hence, the correct option is (B).
- Probability that the man will live for 10 more years = 1/4
- Probability that his wife will live for 10 more years = 1/3
To find: Probability that neither will be alive in 10 years
Solution:
Let A be the event that the man will be alive in 10 years, and B be the event that his wife will be alive in 10 years. Then, the probability that neither will be alive in 10 years is the probability of the complement of the union of A and B, 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 the man and his wife will be alive in 10 years.
Substituting the given probabilities, we get:
P(A ∪ B) = 1/4 + 1/3 - P(A ∩ B)
Simplifying, we get:
P(A ∩ B) = 7/12 - P((A ∪ B)')
Now, we know that the probability of the man being alive in 10 years is 1/4, which means the probability of him not being alive in 10 years is 3/4. Similarly, the probability of his wife not being alive in 10 years is 2/3.
Using the product rule of probability, we can find the probability of both of them not being alive in 10 years:
P((A ∪ B)') = P(A' ∩ B') = P(A') × P(B') = (3/4) × (2/3) = 1/2
Substituting this value in the above equation, we get:
P(A ∩ B) = 7/12 - 1/2 = 1/12
Therefore, the probability that neither the man nor his wife will be alive in 10 years is:
P((A ∪ B)') = 1 - P(A ∪ B) = 1 - (1/4 + 1/3 - 1/12) = 1/2
Hence, the correct option is (B).
A student can solve 70% problems of a book and second student solve 50% problem of same book. Find the probability that at least one of them will solve a selected problem from this book.- a)13/20
- b)17/20
- c)3/20
- d)7/20
Correct answer is option 'B'. Can you explain this answer?
A student can solve 70% problems of a book and second student solve 50% problem of same book. Find the probability that at least one of them will solve a selected problem from this book.
a)
13/20
b)
17/20
c)
3/20
d)
7/20
|
Vijay Kumar answered |
Probability that first and second student can solve
=0.7×0.5=0.35
Probability that first can solve and second cannot solve
=0.7×0.5=0.35
Probability that first cannot solve and Amisha can solve
=0.3×0.5=0.15
Therefore, probability that at least one of them will solve
=0.35+0.35+0.15 = 0.85
=> 85/100
= 17/20
=0.7×0.5=0.35
Probability that first can solve and second cannot solve
=0.7×0.5=0.35
Probability that first cannot solve and Amisha can solve
=0.3×0.5=0.15
Therefore, probability that at least one of them will solve
=0.35+0.35+0.15 = 0.85
=> 85/100
= 17/20
The sample space of an experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH},
then A, B and C form a set of?- a)Mutually exclusive but not exhaustive events
- b)Exhaustive but not mutually exclusive events
- c)Mutually exclusive and exhaustive events
- d)Neither mutually exclusive nor exhaustive events
Correct answer is option 'C'. Can you explain this answer?
The sample space of an experiment is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and A = {TTT}, B = {HTT, THT, TTH}, C = {HHT, HTH, THH, HHH},
then A, B and C form a set of?
then A, B and C form a set of?
a)
Mutually exclusive but not exhaustive events
b)
Exhaustive but not mutually exclusive events
c)
Mutually exclusive and exhaustive events
d)
Neither mutually exclusive nor exhaustive events
|
Suresh Reddy answered |
Above information has no common element in three of them and U of three sets forms sample set . Therefore option C is correct.
Two dice are thrown simultaneously. If X denotes the number of sixes, then the expectation of X is:- a)1/3
- b)1/2
- c)1/4
- d)1/6
Correct answer is option 'A'. Can you explain this answer?
Two dice are thrown simultaneously. If X denotes the number of sixes, then the expectation of X is:
a)
1/3
b)
1/2
c)
1/4
d)
1/6
|
Jayant Mishra answered |
Toolbox:
Mean of the probability distribution = ∑(XixP(Xi))
P (X = 1) = P ( 6 on 1st die and no 6 on 2nd OR 6 on 2nd die and no 6 on 1st)
A die is tossed twice. The probability of getting 1, 2, 3 or 4 on the first toss and 4, 5, or 6 on the second toss is:- a)2/5
- b)1/2
- c)1/3
- d)1/9
Correct answer is option 'C'. Can you explain this answer?
A die is tossed twice. The probability of getting 1, 2, 3 or 4 on the first toss and 4, 5, or 6 on the second toss is:
a)
2/5
b)
1/2
c)
1/3
d)
1/9
|
|
Ritu Singh answered |
In each case, the sample space is given by S={1,2,3,4,5,6}.
Let E = event of getting a 1, 2, 3 or 4 on the first toss.
And, F = event of getting a 5, 6,or 7 on the second toss.
Then, P(E) = 4/6 = 2/3
and P(F) = 3/6 = 1/2
Clearly, E and F are independent events.
∴ required probability = P(E∩F) = P(E)×P(F) [∵ E and F are independent]
= 2/3 * 1/2 = 1/3
Let E = event of getting a 1, 2, 3 or 4 on the first toss.
And, F = event of getting a 5, 6,or 7 on the second toss.
Then, P(E) = 4/6 = 2/3
and P(F) = 3/6 = 1/2
Clearly, E and F are independent events.
∴ required probability = P(E∩F) = P(E)×P(F) [∵ E and F are independent]
= 2/3 * 1/2 = 1/3
If 2/7 is the probability of an event, then the probability of the event ‘not A’ is:- a)5/7
- b)6/7
- c)2/7
- d)3/7
Correct answer is option 'A'. Can you explain this answer?
If 2/7 is the probability of an event, then the probability of the event ‘not A’ is:
a)
5/7
b)
6/7
c)
2/7
d)
3/7
|
|
Geetika Shah answered |
If 2/7 is probability of event A then the probability of an event not A is 1-(2/7) i.e., probability of not A is 5/7
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.- a)7/12
- b)1/4
- c)1/2
- d)3/8
Correct answer is option 'D'. Can you explain this answer?
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.
a)
7/12
b)
1/4
c)
1/2
d)
3/8
|
|
Sarita Yadav answered |
Let E be the event that the man reports that six occurs in the throwing of the die and
let S1 be the event that six occurs and S2 be the event that six does not occur. Then P(S1) = Probability that six occurs = 1/6
P(S2) = Probability that six does not occur = 5/6
P(E|S1) = Probability that the man reports that six occurs when six has actually occurred on the die
= Probability that the man speaks the truth = 3/4
P(E|S2) = Probability that the man reports that six occurs when six has not actually occurred on the die
= Probability that the man does not speak the truth = 1 - 3/4 = 1/4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is actually a six
⇒P(S1/E) = (P(S1)P(E/S1))/(P(S1)P(E/S1) + P(S2)P(E/S2))
= (1/6 x 3/4)/(1/6 x 3/4 + 5/6 x 1/4)
= 1/8 x 24/8
= 3/8
let S1 be the event that six occurs and S2 be the event that six does not occur. Then P(S1) = Probability that six occurs = 1/6
P(S2) = Probability that six does not occur = 5/6
P(E|S1) = Probability that the man reports that six occurs when six has actually occurred on the die
= Probability that the man speaks the truth = 3/4
P(E|S2) = Probability that the man reports that six occurs when six has not actually occurred on the die
= Probability that the man does not speak the truth = 1 - 3/4 = 1/4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is actually a six
⇒P(S1/E) = (P(S1)P(E/S1))/(P(S1)P(E/S1) + P(S2)P(E/S2))
= (1/6 x 3/4)/(1/6 x 3/4 + 5/6 x 1/4)
= 1/8 x 24/8
= 3/8
What is the probability of picking a spade from a normal pack of cards and rolling an odd number on a die?- a)½
- b)0
- c)¼
- d)1/8
Correct answer is option 'D'. Can you explain this answer?
What is the probability of picking a spade from a normal pack of cards and rolling an odd number on a die?
a)
½
b)
0
c)
¼
d)
1/8
|
Alok Mehta answered |
Your hunch is right on track. There are an equal number of odd numbers as there are even numbers painted on a fair die. And there are an equal number of red cards as there are black cards in a standard deck of 52 cards. So the four outcomes you list are equally likely.
More generally:
We can use this same approach in more complicated scenarios, like finding the probability that the number resulting from a toss of a tie is a multiple of 3 and the card drawn from a deck of cards is a queen.
An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is:- a)1/2
- b)3/10
- c)1/10
- d)3/5
Correct answer is option 'A'. Can you explain this answer?
An urn contains five balls. Two balls are drawn and found to be white. The probability that all the balls are white is:
a)
1/2
b)
3/10
c)
1/10
d)
3/5
|
Kiran Mukherjee answered |
Solution:
Given, an urn contains five balls. Let's assume that the five balls are numbered as 1, 2, 3, 4, 5.
P(both balls are white) = P(WW) = (number of ways of selecting 2 white balls)/(number of ways of selecting any 2 balls)
Number of ways of selecting any 2 balls = 5C2 = 10
Number of ways of selecting 2 white balls = 2C2 (both white balls should be selected from 2 white balls) + 3C0 (0 white balls should be selected from 3 black balls) = 1 + 1 = 2
P(WW) = 2/10 = 1/5
Let's consider the following cases:
Case 1: All the balls are white
Number of ways of selecting 2 white balls from 5 white balls = 5C2 = 10
P(all balls are white) = 10/10C2 = 1/10
Case 2: One ball is black and the other ball is white
Number of ways of selecting 1 black ball from 3 black balls and 1 white ball from 2 white balls = 3C1 x 2C1 = 6
P(one ball is black and the other ball is white) = 6/10C2 = 3/5
Therefore, the probability that all the balls are white given that two balls drawn are white is:
P(all balls are white | WW) = P(WW)/[P(WW) + P(one ball is black and the other ball is white)]
= (1/5) / [(1/5) + (3/5)]
= 1/2
Therefore, the correct answer is option 'A'.
Given, an urn contains five balls. Let's assume that the five balls are numbered as 1, 2, 3, 4, 5.
P(both balls are white) = P(WW) = (number of ways of selecting 2 white balls)/(number of ways of selecting any 2 balls)
Number of ways of selecting any 2 balls = 5C2 = 10
Number of ways of selecting 2 white balls = 2C2 (both white balls should be selected from 2 white balls) + 3C0 (0 white balls should be selected from 3 black balls) = 1 + 1 = 2
P(WW) = 2/10 = 1/5
Let's consider the following cases:
Case 1: All the balls are white
Number of ways of selecting 2 white balls from 5 white balls = 5C2 = 10
P(all balls are white) = 10/10C2 = 1/10
Case 2: One ball is black and the other ball is white
Number of ways of selecting 1 black ball from 3 black balls and 1 white ball from 2 white balls = 3C1 x 2C1 = 6
P(one ball is black and the other ball is white) = 6/10C2 = 3/5
Therefore, the probability that all the balls are white given that two balls drawn are white is:
P(all balls are white | WW) = P(WW)/[P(WW) + P(one ball is black and the other ball is white)]
= (1/5) / [(1/5) + (3/5)]
= 1/2
Therefore, the correct answer is option 'A'.
The events when we have no reason to believe that one is more likely to occur than the other is called:- a)Equally likely events
- b)Independent events
- c)Dependent event
- d)Not equally likely events
Correct answer is option 'A'. Can you explain this answer?
The events when we have no reason to believe that one is more likely to occur than the other is called:
a)
Equally likely events
b)
Independent events
c)
Dependent event
d)
Not equally likely events
|
|
Nishtha Sengupta answered |
Equally likely events are events where we have no reason to believe that one event is more likely to occur than the other. In other words, the probability of each event occurring is the same. This means that the outcome of one event does not affect the outcome of the other event. Let's explore this concept further.
Definition of Equally Likely Events:
Equally likely events are events in which each event has the same probability of occurring. For example, when flipping a fair coin, the probability of getting heads is the same as the probability of getting tails. Therefore, in this case, the events "getting heads" and "getting tails" are equally likely.
Characteristics of Equally Likely Events:
- The probability of each event is the same.
- The occurrence of one event does not affect the occurrence of the other event.
- The events are independent of each other.
Example:
Let's consider an example of rolling a fair six-sided die. When rolling the die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Since the die is fair, each outcome has an equal chance of occurring, and therefore, each outcome is equally likely.
Key Points:
- Equally likely events are events where we have no reason to believe that one event is more likely to occur than the other.
- The probability of each event is the same.
- The occurrence of one event does not affect the occurrence of the other event.
- Equally likely events are independent of each other.
Conclusion:
When we have no reason to believe that one event is more likely to occur than the other, we refer to them as equally likely events. These events have the same probability of occurring and are independent of each other. Understanding the concept of equally likely events is important in probability theory and helps in analyzing and predicting outcomes in various situations.
Definition of Equally Likely Events:
Equally likely events are events in which each event has the same probability of occurring. For example, when flipping a fair coin, the probability of getting heads is the same as the probability of getting tails. Therefore, in this case, the events "getting heads" and "getting tails" are equally likely.
Characteristics of Equally Likely Events:
- The probability of each event is the same.
- The occurrence of one event does not affect the occurrence of the other event.
- The events are independent of each other.
Example:
Let's consider an example of rolling a fair six-sided die. When rolling the die, there are six possible outcomes: 1, 2, 3, 4, 5, and 6. Since the die is fair, each outcome has an equal chance of occurring, and therefore, each outcome is equally likely.
Key Points:
- Equally likely events are events where we have no reason to believe that one event is more likely to occur than the other.
- The probability of each event is the same.
- The occurrence of one event does not affect the occurrence of the other event.
- Equally likely events are independent of each other.
Conclusion:
When we have no reason to believe that one event is more likely to occur than the other, we refer to them as equally likely events. These events have the same probability of occurring and are independent of each other. Understanding the concept of equally likely events is important in probability theory and helps in analyzing and predicting outcomes in various situations.
The variance of the distribution is
- a)21/100
- b)30/100
- c)49/100
- d)35/100
Correct answer is option 'A'. Can you explain this answer?
The variance of the distribution is
a)
21/100
b)
30/100
c)
49/100
d)
35/100
|
Poonam Reddy answered |
Mean= E(X) = 
= 0×0.3+1×0.7
=0.7
= 0×0.3+1×0.7
=0.7
= 02×0.3+12×0.7 =0.7
Now, ∴ Var(X)= E(X2)-(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21
Now, ∴ Var(X)= E(X2)-(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21
In a simultaneous toss of two coins, the probability of getting no tail is:- a)1/4
- b)1/2
- c)2
- d)0.1
Correct answer is option 'A'. Can you explain this answer?
In a simultaneous toss of two coins, the probability of getting no tail is:
a)
1/4
b)
1/2
c)
2
d)
0.1
|
Om Desai answered |
Sample space = {HH, HT, TH, TT}
n(SS) = 4
No tail = {HH}
n(No tail) = 1
P(No tail) = n(No tail) / n(SS
= 1/4
n(SS) = 4
No tail = {HH}
n(No tail) = 1
P(No tail) = n(No tail) / n(SS
= 1/4
Let E and F be events of a sample space S of an experiment, then P(E’/F) = …
- a)P(E/F)’
- b)P(E’).P(F)
- c)P(E’)/P(F)
- d)1 – P(E/F)
Correct answer is option 'D'. Can you explain this answer?
Let E and F be events of a sample space S of an experiment, then P(E’/F) = …
a)
P(E/F)’
b)
P(E’).P(F)
c)
P(E’)/P(F)
d)
1 – P(E/F)
|
|
Nitin Das answered |
Conditional Probability:
Conditional probability is the probability of an event occurring given that another event has occurred. It is denoted by P(E/F), which means the probability of event E given that event F has occurred.
Formula:
The formula for conditional probability is as follows:
P(E/F) = P(E ∩ F) / P(F)
Here, P(E ∩ F) represents the probability of both events E and F occurring together.
Solution:
Given, E and F are two events of a sample space S.
We need to find the value of P(E/F).
The correct option is D, which means 1.
Explanation:
When we say P(E/F), it means the probability of event E given that event F has occurred.
If the probability of event F happening is 0, then P(E/F) is undefined.
If the probability of event F happening is 1, then P(E/F) is equal to P(E), which means the probability of event E happening.
Here, the given option D is 1, which means that event F has occurred for sure.
So, the probability of event E happening given that event F has occurred is 1, which means that event E has also occurred for sure.
Hence, the correct option is D, which means 1.
Conditional probability is the probability of an event occurring given that another event has occurred. It is denoted by P(E/F), which means the probability of event E given that event F has occurred.
Formula:
The formula for conditional probability is as follows:
P(E/F) = P(E ∩ F) / P(F)
Here, P(E ∩ F) represents the probability of both events E and F occurring together.
Solution:
Given, E and F are two events of a sample space S.
We need to find the value of P(E/F).
The correct option is D, which means 1.
Explanation:
When we say P(E/F), it means the probability of event E given that event F has occurred.
If the probability of event F happening is 0, then P(E/F) is undefined.
If the probability of event F happening is 1, then P(E/F) is equal to P(E), which means the probability of event E happening.
Here, the given option D is 1, which means that event F has occurred for sure.
So, the probability of event E happening given that event F has occurred is 1, which means that event E has also occurred for sure.
Hence, the correct option is D, which means 1.
Probability that A speaks truth is 5/9 . A coin is tossed and reports that a head appears. The probability that actually there was head is:- a)5/9
- b)5/18
- c)2/9
- d)4/9
Correct answer is option 'A'. Can you explain this answer?
Probability that A speaks truth is 5/9 . A coin is tossed and reports that a head appears. The probability that actually there was head is:
a)
5/9
b)
5/18
c)
2/9
d)
4/9
|
|
Sarita Yadav answered |
Let E : A speaks truth
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9)
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth = ½
P(E/H) = Probability that appears head, if A speaks lies = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9
Let E : A speaks truth
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9)
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth = ½
P(E/H) = Probability that appears head, if A speaks lies = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9)
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth = ½
P(E/H) = Probability that appears head, if A speaks lies = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9
Let E : A speaks truth
F : A lies
H : head appears on the toss of coin
P(E) = Probability A speaks truth
= 5/9
P(F) = Probability A speaks lies
1 - P(E) => 1-(5/9)
=> 4/9
P(H/E) = Probability that appears head, if A speaks truth = ½
P(E/H) = Probability that appears head, if A speaks lies = 1/2
P(H/E) = [(5/9) (½)]/[(5/9) (½) + (4/9) (½)]
= (5/18)/[(5/18) + (4/18)]
= 5/9
If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(B|A) = 0.6,then P(A|B) = ……- a)0.1
- b)0.5
- c)0.36
- d)0.2
Correct answer is option 'D'. Can you explain this answer?
If A and B are two events such that P(A) = 0.3 and P(B) = 0.9 and P(B|A) = 0.6,then P(A|B) = ……
a)
0.1
b)
0.5
c)
0.36
d)
0.2
|
Poojitha Seetha answered |
Given, P(A)=0.3, P(B)=0.9, P(B|A)=0.6,
We know that P(B|A)=P(A ∩ B)/P(A) implies P(A ∩ B)=P(B|A).P(A)
Therefore, P(A ∩ B)=0.6*0.3=0.18. Now P(A|B)=P(A ∩ B)/P(B) implies P(A|B)=0.18/0.9=0.2. So, Correct option is 'D' .
Two students Amit and Priyanka appeared in an examination. The probability that Amit will qualify the examination is 0.06 and that Priyanka will qualify the examination is 0.12. The probability that both will qualify the examination is 0.02. Then, the probability that both Amit and Priyanka will not qualify the examination is:- a)0.91
- b)0.12
- c)0.84
- d)0.45
Correct answer is option 'C'. Can you explain this answer?
Two students Amit and Priyanka appeared in an examination. The probability that Amit will qualify the examination is 0.06 and that Priyanka will qualify the examination is 0.12. The probability that both will qualify the examination is 0.02. Then, the probability that both Amit and Priyanka will not qualify the examination is:
a)
0.91
b)
0.12
c)
0.84
d)
0.45
|
|
Naina Sharma answered |
Probability that Amit will qualify the exam(E) = 0.06
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F) - P(EUF)
= 0.06 + 0.12 - 0.02 = 0.16
P(E’⋂F’) = 1 - P(EUF)
⇒ 1 - 0.16
= 0.84
Probability that Priyanka will qualify the exam(F) = 0.12
Probability that both will qualify the exam P(E⋂F) = 0.02
P(EUF) = P(E) + P(F) - P(EUF)
= 0.06 + 0.12 - 0.02 = 0.16
P(E’⋂F’) = 1 - P(EUF)
⇒ 1 - 0.16
= 0.84
If an event has more than one sample point, it is called a ………- a)Exhaustive event
- b)Compound event
- c)Simple event
- d)Mutually exclusive event
Correct answer is option 'B'. Can you explain this answer?
If an event has more than one sample point, it is called a ………
a)
Exhaustive event
b)
Compound event
c)
Simple event
d)
Mutually exclusive event
|
|
Kritika Khanna answered |
Compound Event Explanation:
Compound events are events that have more than one sample point. These events involve multiple outcomes or scenarios, making them more complex than simple events.
Characteristics of Compound Events:
- Compound events consist of multiple sample points or outcomes.
- They can be composed of a combination of simple events.
- Compound events are often more intricate to analyze and calculate compared to simple events.
Example of Compound Event:
For example, rolling a dice and flipping a coin at the same time would create a compound event. The possible outcomes would be a combination of the results from rolling the dice (1-6) and flipping the coin (heads or tails).
Significance of Compound Events:
Understanding compound events is crucial in probability theory and statistics. By analyzing compound events, one can calculate the likelihood of various outcomes occurring simultaneously. This knowledge is valuable in making informed decisions and predictions in various fields such as finance, sports, and scientific research.
In conclusion, compound events play a significant role in probability theory as they involve multiple sample points or outcomes, making them more complex and challenging to analyze compared to simple events.
Compound events are events that have more than one sample point. These events involve multiple outcomes or scenarios, making them more complex than simple events.
Characteristics of Compound Events:
- Compound events consist of multiple sample points or outcomes.
- They can be composed of a combination of simple events.
- Compound events are often more intricate to analyze and calculate compared to simple events.
Example of Compound Event:
For example, rolling a dice and flipping a coin at the same time would create a compound event. The possible outcomes would be a combination of the results from rolling the dice (1-6) and flipping the coin (heads or tails).
Significance of Compound Events:
Understanding compound events is crucial in probability theory and statistics. By analyzing compound events, one can calculate the likelihood of various outcomes occurring simultaneously. This knowledge is valuable in making informed decisions and predictions in various fields such as finance, sports, and scientific research.
In conclusion, compound events play a significant role in probability theory as they involve multiple sample points or outcomes, making them more complex and challenging to analyze compared to simple events.
A fair six-sided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?- a)1/36
- b)1/18
- c)5/36
- d)1/6
- e)1/3
Correct answer is option 'C'. Can you explain this answer?
A fair six-sided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?
a)
1/36
b)
1/18
c)
5/36
d)
1/6
e)
1/3
|
Nabanita Basu answered |
The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.
P(getting first 2) = 1/6
P(no second 4) = 5/6
Therefore P(getting first 2 and no second 4) = 1/6* 5/6 = 5/36
Five marbles are drawn from a bag which contains 6 blue marbles and 7 green marbles. Then, the probability that 3 will be blue and 2 green is:- a)213/429
- b)140/429
- c)117/429
- d)167/429
Correct answer is option 'B'. Can you explain this answer?
Five marbles are drawn from a bag which contains 6 blue marbles and 7 green marbles. Then, the probability that 3 will be blue and 2 green is:
a)
213/429
b)
140/429
c)
117/429
d)
167/429
|
|
Naina Sharma answered |
Blue = 6
Green = 7
Number of ways 3 blue marbles can be drawn = 6C3
number of ways 2 green marbles can be drawn = 7C2
total ways of drawing 5 marbles = 13C5
=> probability = (6C3 * 7C2)/13C5
= 140/429
Green = 7
Number of ways 3 blue marbles can be drawn = 6C3
number of ways 2 green marbles can be drawn = 7C2
total ways of drawing 5 marbles = 13C5
=> probability = (6C3 * 7C2)/13C5
= 140/429
Two parts A and B of a machine is manufactured by a firm. Out of 100 A’s 12 are likely to be defective and Out of 100 B’s 8 are likely to be defective. The probability that a machine manufactured by the firm is free from any defect is:- a)619/625
- b)119/625
- c)6/625
- d)506/625
Correct answer is option 'D'. Can you explain this answer?
Two parts A and B of a machine is manufactured by a firm. Out of 100 A’s 12 are likely to be defective and Out of 100 B’s 8 are likely to be defective. The probability that a machine manufactured by the firm is free from any defect is:
a)
619/625
b)
119/625
c)
6/625
d)
506/625
|
|
Aryan Khanna answered |
probability of getting good machine part from A = 88/100
probability of getting good machine part from B = 92/100
P(correction) = (88/100)(92/100)
= 506/625
probability of getting good machine part from B = 92/100
P(correction) = (88/100)(92/100)
= 506/625
If A and B are independent events, such that P(A ∪ B)= 0.7, P(B) = 0.5, then P(A) = ……- a)0.2
- b)0.3
- c)0.4
- d)0.6
Correct answer is option 'C'. Can you explain this answer?
If A and B are independent events, such that P(A ∪ B)= 0.7, P(B) = 0.5, then P(A) = ……
a)
0.2
b)
0.3
c)
0.4
d)
0.6
|
|
Nikita Singh answered |
P(A ∪ B) = P(A) + P(B) - P(A)P(B)
= 0.7 = P(A) + 0.5 - 0.5 P(A)
0.2 = 0.5 P(A)
P(A) = ⅖
P(A) = 0.4
= 0.7 = P(A) + 0.5 - 0.5 P(A)
0.2 = 0.5 P(A)
P(A) = ⅖
P(A) = 0.4
The letters of the word ‘ASSASSIN ‘ are written at random in a row. The chance that all the similar letters occur together is- a)1/35
- b)1/14
- c)34/35
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The letters of the word ‘ASSASSIN ‘ are written at random in a row. The chance that all the similar letters occur together is
a)
1/35
b)
1/14
c)
34/35
d)
none of these
|
|
Ankita Chavan answered |
There is no specific word mentioned in your question. Please provide the word you are referring to.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.- a)47/91
- b)49/91
- c)4191
- d)44/91
Correct answer is option 'D'. Can you explain this answer?
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
a)
47/91
b)
49/91
c)
4191
d)
44/91
|
Anshika Rane answered |
Total oranges = 15., Good Oranges = 12. Let G stands for a good orange. Therefore , Required Probability = P(GGG) = P(G).P(G/G).P(G/GG) 
If E and F are events then P (E ∩ F) =- a)P (E ∩ F) P (F|E), P (E) ≠ 0
- b)P (E) P (E|F), P (E) ≠ 0
- c)P (E∪F) P (F|E), P (E) ≠ 0
- d)P (E) P (F|E), P (E) ≠ 0
Correct answer is option 'D'. Can you explain this answer?
If E and F are events then P (E ∩ F) =
a)
P (E ∩ F) P (F|E), P (E) ≠ 0
b)
P (E) P (E|F), P (E) ≠ 0
c)
P (E∪F) P (F|E), P (E) ≠ 0
d)
P (E) P (F|E), P (E) ≠ 0
|
Athira Datta answered |
If E and F are events then P (E ∩ F) = P (E) P (F|E), P (E) ≠ 0.
8 coins are tossed at a time. The probability of getting 6 heads up is- a)57/64
- b)7/64
- c)37/256
- d)229/256
Correct answer is option 'C'. Can you explain this answer?
8 coins are tossed at a time. The probability of getting 6 heads up is
a)
57/64
b)
7/64
c)
37/256
d)
229/256
|
|
Koyna Jatwar answered |
The required probability =8C6(1/2)^6.(1/2)^2+8C7(1/2)^7(1/2)+8C8(1/2)^8. =1/256(28+8+1)=37/256
An urn contains 3 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. The possible values of X are:- a)1 & 2
- b)0, 1 & 2
- c)0, 1, 2 & 3
- d)1, 2 & 3
Correct answer is option 'B'. Can you explain this answer?
An urn contains 3 red and 2 black balls. Two balls are randomly drawn. Let X represents the number of black balls. The possible values of X are:
a)
1 & 2
b)
0, 1 & 2
c)
0, 1, 2 & 3
d)
1, 2 & 3
|
|
Rajat Patel answered |
Given that an urn has 3 Red balls and 2 Black balls.
The number of ways in which two balls can be reprsented are {(RR), (RB), (BR), (BB)}
Let X represent the number of black balls. Possible values of X are: X (RR) = 0, X (RB) = 1, X(BR = 1) and X (BB) = 2.
Therefore the possible values of X are 0, 1 and 2.
Random variable is a real valued function whose domain is the sample space of a ……… and range is the set of ………
- a) Random variable, non-negative integers
- b) Random variable, real numbers
- c) Real numbers, real numbers
- d) Random experiment, real numbers
Correct answer is option 'D'. Can you explain this answer?
Random variable is a real valued function whose domain is the sample space of a ……… and range is the set of ………
a)
Random variable, non-negative integersb)
Random variable, real numbersc)
Real numbers, real numbersd)
Random experiment, real numbers|
|
Kritika Khanna answered |
The correct answer is option D: Random experiment, real numbers.
Explanation:
A random variable is a real-valued function that maps each outcome of a random experiment to a real number. To understand why the correct answer is option D, let's break down the terms involved in the question.
1. Random variable:
A random variable is a variable whose value is determined by the outcome of a random experiment. It is denoted by a capital letter like X. The random variable assigns a numerical value to each possible outcome of the experiment.
2. Random experiment:
A random experiment is a process or procedure that generates a set of outcomes. It is characterized by its sample space, which is the set of all possible outcomes of the experiment. Examples of random experiments include tossing a coin, rolling a die, or selecting a card from a deck.
3. Domain and range:
The domain of a function is the set of all possible inputs or values that the function can take. In the context of a random variable, the domain is the sample space of the random experiment, as the values of the random variable are determined by the outcomes of the experiment.
The range of a function is the set of all possible outputs or values that the function can produce. In the case of a random variable, the range is the set of real numbers. This is because the random variable assigns a real number to each outcome of the random experiment.
Therefore, the correct answer is option D: Random experiment, real numbers. The domain of the random variable is the sample space of the random experiment, and the range is the set of real numbers.
Explanation:
A random variable is a real-valued function that maps each outcome of a random experiment to a real number. To understand why the correct answer is option D, let's break down the terms involved in the question.
1. Random variable:
A random variable is a variable whose value is determined by the outcome of a random experiment. It is denoted by a capital letter like X. The random variable assigns a numerical value to each possible outcome of the experiment.
2. Random experiment:
A random experiment is a process or procedure that generates a set of outcomes. It is characterized by its sample space, which is the set of all possible outcomes of the experiment. Examples of random experiments include tossing a coin, rolling a die, or selecting a card from a deck.
3. Domain and range:
The domain of a function is the set of all possible inputs or values that the function can take. In the context of a random variable, the domain is the sample space of the random experiment, as the values of the random variable are determined by the outcomes of the experiment.
The range of a function is the set of all possible outputs or values that the function can produce. In the case of a random variable, the range is the set of real numbers. This is because the random variable assigns a real number to each outcome of the random experiment.
Therefore, the correct answer is option D: Random experiment, real numbers. The domain of the random variable is the sample space of the random experiment, and the range is the set of real numbers.
The letters of the word ‘MALENKOV‘are arranged in all possible ways. The chance that there are exactly four letters between M and E is- a)3/28
- b)3/14
- c)1/14
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The letters of the word ‘MALENKOV‘are arranged in all possible ways. The chance that there are exactly four letters between M and E is
a)
3/28
b)
3/14
c)
1/14
d)
none of these
|
|
Shilpa Rane answered |
"algorithm" can be rearranged in 9! (362,880) different ways.
What is the sample space for an experiment when a coin is tossed and then a dice is thrown?- a){1H,2H3H,4H,5H,6H,T1,T2,T3,T4,T5,T6}
- b){H1,T1,H6,T6}
- c){H1,T1,H2,T2,H3,T3,H4,T4,H5,T5,H6,T6}
- d){HT,TH,TT,HH,1,2,3,4,5,6}
Correct answer is option 'C'. Can you explain this answer?
What is the sample space for an experiment when a coin is tossed and then a dice is thrown?
a)
{1H,2H3H,4H,5H,6H,T1,T2,T3,T4,T5,T6}
b)
{H1,T1,H6,T6}
c)
{H1,T1,H2,T2,H3,T3,H4,T4,H5,T5,H6,T6}
d)
{HT,TH,TT,HH,1,2,3,4,5,6}
|
|
Nandini Iyer answered |
When a coin is tossed.Head or Tail may occur.
Whereby when a die is thrown the numbers from 1 to 6 may occur.
H represents Head.
T represents Tail.
(1-6) sides of the die.
Which of the following conditions do Bernoulli trials satisfy?- a)Each trial has finite number of outcomes and the probability of each outcome remains the same in each trial.
- b)Each trial has exactly two outcomes: success or failure and the probability of success or failure can vary in each trial.
- c)Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.
- d)Each trial has exactly three outcomes: success or failure and the probability of success or failure remains the same in each trial.
Correct answer is option 'C'. Can you explain this answer?
Which of the following conditions do Bernoulli trials satisfy?
a)
Each trial has finite number of outcomes and the probability of each outcome remains the same in each trial.
b)
Each trial has exactly two outcomes: success or failure and the probability of success or failure can vary in each trial.
c)
Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.
d)
Each trial has exactly three outcomes: success or failure and the probability of success or failure remains the same in each trial.
|
|
Nabanita Basu answered |
Bernoulli trials are true only when Each trial has exactly two outcomes: success or failure and the probability of success or failure remains the same in each trial.
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.- a)27/102
- b)29/102
- c)25/102
- d)23/102
Correct answer is option 'C'. Can you explain this answer?
Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
a)
27/102
b)
29/102
c)
25/102
d)
23/102
|
Anjali Deshpande answered |
Required probability : P(BB) = P(B) X P(B/B) ……………{B means black card}
In a meeting, 60% of the members favour and 40% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).- a)0.6 and 0
- b)0.6 and 0.44
- c)0.6 and 0.24
- d)0.4 and 0.24
Correct answer is option 'C'. Can you explain this answer?
In a meeting, 60% of the members favour and 40% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var(X).
a)
0.6 and 0
b)
0.6 and 0.44
c)
0.6 and 0.24
d)
0.4 and 0.24
|
|
Vaishnavi Iyer answered |
Solution:
Given,
P(X=1) = 0.6, P(X=0) = 0.4
We know that,
E(X) = ΣxP(x)
So,
E(X) = (0 × 0.4) + (1 × 0.6)
E(X) = 0.6
Now, we need to find the variance of X, which can be calculated as follows:
Var(X) = E(X^2) - [E(X)]^2
We know that,
E(X^2) = Σx^2P(x)
So,
E(X^2) = (0^2 × 0.4) + (1^2 × 0.6)
E(X^2) = 0.6
Therefore,
Var(X) = 0.6 - (0.6)^2
Var(X) = 0.24
Hence, the correct answer is option C, i.e., 0.6 and 0.24.
Given,
P(X=1) = 0.6, P(X=0) = 0.4
We know that,
E(X) = ΣxP(x)
So,
E(X) = (0 × 0.4) + (1 × 0.6)
E(X) = 0.6
Now, we need to find the variance of X, which can be calculated as follows:
Var(X) = E(X^2) - [E(X)]^2
We know that,
E(X^2) = Σx^2P(x)
So,
E(X^2) = (0^2 × 0.4) + (1^2 × 0.6)
E(X^2) = 0.6
Therefore,
Var(X) = 0.6 - (0.6)^2
Var(X) = 0.24
Hence, the correct answer is option C, i.e., 0.6 and 0.24.
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is- a)55/2048
- b)3/4096
- c)1/1024
- d)3/1024
Correct answer is option 'A'. Can you explain this answer?
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is
a)
55/2048
b)
3/4096
c)
1/1024
d)
3/1024
|
Sanchita Khanna answered |
Given, probability of getting 4 heads = probability of getting 7 heads
Let the number of tosses be n. Then,
Probability of getting 4 heads = Probability of getting (n-4) tails = (1/2)^n * nC4
Probability of getting 7 heads = Probability of getting (n-7) tails = (1/2)^n * nC7
Equating the two probabilities, we get
nC4 = nC7/35
Solving this equation, we get n = 11
So, the number of tosses is 11.
Now, we need to find the probability of getting 2 heads.
Probability of getting 2 heads = Probability of getting (11-2) tails = (1/2)^11 * nC2
= (1/2)^11 * 55
= 55/2048
Therefore, the correct option is A) 55/2048.
Let the number of tosses be n. Then,
Probability of getting 4 heads = Probability of getting (n-4) tails = (1/2)^n * nC4
Probability of getting 7 heads = Probability of getting (n-7) tails = (1/2)^n * nC7
Equating the two probabilities, we get
nC4 = nC7/35
Solving this equation, we get n = 11
So, the number of tosses is 11.
Now, we need to find the probability of getting 2 heads.
Probability of getting 2 heads = Probability of getting (11-2) tails = (1/2)^11 * nC2
= (1/2)^11 * 55
= 55/2048
Therefore, the correct option is A) 55/2048.
From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards is- a)109/153
- b)23/27
- c)46/153
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
From a pack of 52 cards, the cards are drawn one by one till an ace appears. The chance that an ace does not come up in first 26 cards is
a)
109/153
b)
23/27
c)
46/153
d)
none of these
|
|
Madhavan Chatterjee answered |
There are 4 aces in a pack. So, the probability of drawing an ace = 4/52
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
= 1/13
and the probability of drawing a card other than ace = 12/13.
Hence, the probability that ace doesn't appear in first 26 draws = (12/13)26
Ashmit can solve 80% of the problem given in a book and Amisha can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?
- a)0.60
- b)0.06
- c)0.94
- d)0.56
Correct answer is option 'C'. Can you explain this answer?
Ashmit can solve 80% of the problem given in a book and Amisha can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?
a)
0.60
b)
0.06
c)
0.94
d)
0.56
|
|
Rounak Verma answered |
Probability that both Ashmit and Amisha can solve
=0.8×0.7=0.56
Probability that Ashmit can solve but Amisha not
=0.8×0.3=0.24
Probability that Amisha can solve but Ashmit not
=0.2×0.7=0.14
so atleast one of them solve the problem
= 0.56+0.24+0.14
=0.94
=0.8×0.7=0.56
Probability that Ashmit can solve but Amisha not
=0.8×0.3=0.24
Probability that Amisha can solve but Ashmit not
=0.2×0.7=0.14
so atleast one of them solve the problem
= 0.56+0.24+0.14
=0.94
The probability that a card drawn at random from a pack of 52 cards is a king or a heart is- a)16/52
- b)1/13
- c)1/52
- d)1/4
Correct answer is option 'A'. Can you explain this answer?
The probability that a card drawn at random from a pack of 52 cards is a king or a heart is
a)
16/52
b)
1/13
c)
1/52
d)
1/4
|
Er Aarif answered |
The number of kings in a deck are 4
Number of hearts in the deck are 13, including the king of hearts
Probability of getting either a king or a heart is, P = 4+(13–1) / 52
P = 16/52
Number of hearts in the deck are 13, including the king of hearts
Probability of getting either a king or a heart is, P = 4+(13–1) / 52
P = 16/52
If A and B are two events of sample space S, then- a)P(A ∩ B) = P(B)P(A/B); P(B) ≠ 0
- b)P(A ∪ B) = P(A)P(A/B); P(B) ≠ 0
- c)P(A ∪ B) = P(B)P(A/B); P(B) ≠ 0
- d)P(A ∩ B) = P(A)P(A/B); P(B) ≠ 0
Correct answer is option 'A'. Can you explain this answer?
If A and B are two events of sample space S, then
a)
P(A ∩ B) = P(B)P(A/B); P(B) ≠ 0
b)
P(A ∪ B) = P(A)P(A/B); P(B) ≠ 0
c)
P(A ∪ B) = P(B)P(A/B); P(B) ≠ 0
d)
P(A ∩ B) = P(A)P(A/B); P(B) ≠ 0
|
|
Poojitha Seetha answered |
The probability of occurrence of event A under the condition that event B has already occurred& P(B)≠0 is called Conditional probability i.e; P(A|B)=P(A ∩ B)/P(B). Multiply with P(B) on both sides implies P(A ∩ B)=P(B).P(A|B). So option 'A' is correct.
Chapter doubts & questions for Unit 4: Probability, Random Variables, and Probability Distributions - AP Statistics 2024 is part of Grade 9 exam preparation. The chapters have been prepared according to the Grade 9 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Grade 9 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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