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All questions of Probability Class 12 for JEE Exam
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 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
|
Infinity Academy answered |
There are 12 even numbers between 1 to 25
Consider the given events.
A = Even number ticket in the first draw
B = Even number ticket in the second draw
Now, P(A) = 12/25
P(B/A) = 11/24
Required probability : P(A⋂B) = P(A) * P(B/A)
= (12/25) * (11/24)
= 11/50
Consider the given events.
A = Even number ticket in the first draw
B = Even number ticket in the second draw
Now, P(A) = 12/25
P(B/A) = 11/24
Required probability : P(A⋂B) = P(A) * P(B/A)
= (12/25) * (11/24)
= 11/50
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 six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is (2005S)- a)5/11
- b)5/6
- c)6/11
- d)1/6
Correct answer is option 'A'. Can you explain this answer?
A six faced fair dice is thrown until 1 comes, then the probability that 1 comes in even no. of trials is (2005S)
a)
5/11
b)
5/6
c)
6/11
d)
1/6
|
Krishna Iyer answered |
In single throw of a dice, probability of getting 1 is = 
and prob. of not getting 1 is 
and prob. of not getting 1 is Then getting 1 in even no. of chances = getting 1 in 2nd chance or in 4th chance or in 6th chance and so on
∴ Req. Prob 
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
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)
Two cards are drawn at random from a pack of playing cards. Find the probability that one card is a heart and the other is an ace.[REE 2001 (Mains)]- a)2/26
- b)1/26
- c)3/26
- d)4/26
Correct answer is option 'B'. Can you explain this answer?
Two cards are drawn at random from a pack of playing cards. Find the probability that one card is a heart and the other is an ace.
[REE 2001 (Mains)]
a)
2/26
b)
1/26
c)
3/26
d)
4/26
|
Lavanya Menon answered |
1/26 is the probability of getting a heart or an ace when drawing a single card from a deck of 52 cards.
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
A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is:- a)4/9
- b)2/9
- c)5/9
- d)1/3
Correct answer is option 'A'. Can you explain this answer?
A die is tossed twice. Getting a number greater than 4 is considered a success. Then the variance of the probability distribution of the number of successes is:
a)
4/9
b)
2/9
c)
5/9
d)
1/3
|
|
Praveen Kumar answered |
In a single toss, P(success) = 2/6 = 1/3 and P(non-success) = (1 − 1/3) = 2/3.
P(X=0) = P(non-success in the 1st draw and non-success in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and non-success in the 2nd) or (non-success in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.
P(X=0) = P(non-success in the 1st draw and non-success in the second)
(2/3 × 2/3) = 4/9.
P(X=1) = P(success in the 1st toss and non-success in the 2nd) or (non-success in the 1st toss and success in the 2nd)]
(1/3 x 2/3) + (2/3 × 1/3) = 4/9.
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.
A random variable X has the following probability distribution: 
Then k =?- a)1/8
- b)8
- c)0
- d)4
Correct answer is option 'A'. Can you explain this answer?
A random variable X has the following probability distribution:
Then k =?
a)
1/8
b)
8
c)
0
d)
4
|
Chan singh answered |
We know that P(X)=1 ie
P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1
0+k+2k+2k+3k=1
k= ⅛
P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=1
0+k+2k+2k+3k=1
k= ⅛
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 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
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is-[AIEEE 2003]- a)2/5
- b)4/5
- c)3/5
- d)1/5
Correct answer is option 'A'. Can you explain this answer?
Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is-
[AIEEE 2003]
a)
2/5
b)
4/5
c)
3/5
d)
1/5
|
Ruchi Yadav answered |
Let H1H2H3H4 B be the horses in which B is the winning horse.
Required probability
Required probability
An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is then: (1993 - 1 Mark)- a)16 / 81
- b)1 / 81
- c)80 / 81
- d)65 / 81
Correct answer is option 'A'. Can you explain this answer?
An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5, is then: (1993 - 1 Mark)
a)
16 / 81
b)
1 / 81
c)
80 / 81
d)
65 / 81
|
Ankita Das answered |
Option a is correct
Three identical dice are rolled. The probability that the same number will appear on each of them is (1984 - 2 Marks)- a)1/6
- b)1/36
- c)1/18
- d)3/28
Correct answer is option 'B'. Can you explain this answer?
Three identical dice are rolled. The probability that the same number will appear on each of them is (1984 - 2 Marks)
a)
1/6
b)
1/36
c)
1/18
d)
3/28
|
|
Ananya Das answered |
Favourable cases = 6; {(1, 1, 1), (2, 2, 2),…(6, 6, 6)}
Total cases = 6 × 6 × 6 = 216
∴Req. prob. =
∴Req. prob. =
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' .
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
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
A random variable X has the probabil ity distribution : 
For the events E = {X is a prime number} and F = {X < 4}, the probability P(E ∪ F) is[AIEEE 2004]- a)0.87
- b)0.77
- c)0.35
- d)0.50
Correct answer is option 'B'. Can you explain this answer?
A random variable X has the probabil ity distribution : For the events E = {X is a prime number} and F = {X < 4}, the probability P(E ∪ F) is
For the events E = {X is a prime number} and F = {X < 4}, the probability P(E ∪ F) is[AIEEE 2004]
a)
0.87
b)
0.77
c)
0.35
d)
0.50
|
Vp Classes answered |
P (E) = P (2 or 3 or 5 or 7) = 0.23 + 0.12 + 0.20 + 0.07
= 0.6
P (F) = P (1 or 2 or 3)
= 0.15 + 0.23 + 0.12
= 0.50
P (E ⋂ F) = P (2 OR 3)=0.23+0.12=0.35
P (E ⋂ F) = P (E) + P (F) - P (E U F)
0.35 = 0.62 + 0.50 - P (E U F)
= 0.77
= 0.6
P (F) = P (1 or 2 or 3)
= 0.15 + 0.23 + 0.12
= 0.50
P (E ⋂ F) = P (2 OR 3)=0.23+0.12=0.35
P (E ⋂ F) = P (E) + P (F) - P (E U F)
0.35 = 0.62 + 0.50 - P (E U F)
= 0.77
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
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
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.
A fair die is tossed repeatedly until a six is obtained. Let X denA fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The probability that X > 3 equals[JEE 2009]- a)125/216
- b)25/36
- c)5/36
- d)25/216
Correct answer is option 'B'. Can you explain this answer?
A fair die is tossed repeatedly until a six is obtained. Let X denA fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required. The probability that X > 3 equals
[JEE 2009]
a)
125/216
b)
25/36
c)
5/36
d)
25/216
|
Sahana Ahuja answered |
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.
A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A È B) is[AIEEE 2008]- a)0
- b)1
- c)2/5
- d)3/5
Correct answer is option 'B'. Can you explain this answer?
A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A È B) is
[AIEEE 2008]
a)
0
b)
1
c)
2/5
d)
3/5
|
Arka Chavan answered |
Probability of Events A and B
-------------------------------------
Let's consider a standard six-sided die. When the die is thrown, there are six possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6.
Event A: The number obtained is greater than 3. This means that Event A consists of the outcomes 4, 5, and 6.
Event B: The number obtained is less than 5. This means that Event B consists of the outcomes 1, 2, 3, and 4.
To find the probability of Event A and Event B occurring simultaneously, we need to find the intersection of Event A and Event B, denoted as A ∩ B.
Intersection of Events A and B
--------------------------------
The intersection of two events A and B represents the outcomes that are common to both A and B. In this case, the outcomes 4 and 5 are common to both Event A and Event B.
So, A ∩ B = {4, 5}.
Probability of A ∩ B
----------------------
The probability of A ∩ B can be calculated using the formula:
P(A ∩ B) = P(A) × P(B|A)
P(A) represents the probability of Event A occurring, and P(B|A) represents the probability of Event B occurring given that Event A has already occurred.
Since A ∩ B consists of two outcomes (4 and 5) and there are six possible outcomes in total, the probability of A ∩ B is:
P(A ∩ B) = 2/6 = 1/3
So, the correct answer is option B) 1.
-------------------------------------
Let's consider a standard six-sided die. When the die is thrown, there are six possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6.
Event A: The number obtained is greater than 3. This means that Event A consists of the outcomes 4, 5, and 6.
Event B: The number obtained is less than 5. This means that Event B consists of the outcomes 1, 2, 3, and 4.
To find the probability of Event A and Event B occurring simultaneously, we need to find the intersection of Event A and Event B, denoted as A ∩ B.
Intersection of Events A and B
--------------------------------
The intersection of two events A and B represents the outcomes that are common to both A and B. In this case, the outcomes 4 and 5 are common to both Event A and Event B.
So, A ∩ B = {4, 5}.
Probability of A ∩ B
----------------------
The probability of A ∩ B can be calculated using the formula:
P(A ∩ B) = P(A) × P(B|A)
P(A) represents the probability of Event A occurring, and P(B|A) represents the probability of Event B occurring given that Event A has already occurred.
Since A ∩ B consists of two outcomes (4 and 5) and there are six possible outcomes in total, the probability of A ∩ B is:
P(A ∩ B) = 2/6 = 1/3
So, the correct answer is option B) 1.
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.
Four cards are drawn from a pack of 52 playing cards. Find the probability of drawing exactly one pair. [REE 99, 6]
Correct answer is '0.304'. Can you explain this answer?
Four cards are drawn from a pack of 52 playing cards. Find the probability of drawing exactly one pair.
[REE 99, 6]
|
|
Anirban Desai answered |
Calculation of Probability of Drawing Exactly One Pair
To find the probability of drawing exactly one pair from a deck of 52 playing cards, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
Favorable Outcomes:
To have exactly one pair, we need to select two cards of the same rank and two cards of different ranks. Let's break down the calculation of favorable outcomes into two steps:
Step 1: Selecting a rank for the pair
- There are 13 ranks in a deck of playing cards (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K).
- We need to choose 1 rank out of these 13 ranks, which can be done in 13 ways.
Step 2: Selecting the cards for the pair and the remaining two cards
- Once we have selected a rank for the pair, we need to choose 2 cards of that rank from the deck, which can be done in (4 choose 2) = 6 ways.
- The remaining two cards should be of different ranks, so we have 12 ranks left to choose from.
- We need to choose 1 card of one rank and 1 card of another rank, which can be done in (4 choose 1) * (4 choose 1) = 16 ways.
Therefore, the total number of favorable outcomes is 13 * 6 * 16 = 1248.
Total Number of Possible Outcomes:
To calculate the total number of possible outcomes, we need to choose 4 cards from a deck of 52 cards, which can be done in (52 choose 4) = 270,725 ways.
Calculation of Probability:
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Therefore, the probability of drawing exactly one pair can be calculated as follows:
Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes
= 1248 / 270,725
≈ 0.0046
However, the given correct answer is '0.304', which means the probability needs to be expressed as a percentage. To convert the probability to a percentage, we multiply it by 100.
Probability (as a percentage) = 0.0046 * 100
≈ 0.46%
Therefore, the probability of drawing exactly one pair is approximately 0.46%.
To find the probability of drawing exactly one pair from a deck of 52 playing cards, we need to calculate the number of favorable outcomes and the total number of possible outcomes.
Favorable Outcomes:
To have exactly one pair, we need to select two cards of the same rank and two cards of different ranks. Let's break down the calculation of favorable outcomes into two steps:
Step 1: Selecting a rank for the pair
- There are 13 ranks in a deck of playing cards (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K).
- We need to choose 1 rank out of these 13 ranks, which can be done in 13 ways.
Step 2: Selecting the cards for the pair and the remaining two cards
- Once we have selected a rank for the pair, we need to choose 2 cards of that rank from the deck, which can be done in (4 choose 2) = 6 ways.
- The remaining two cards should be of different ranks, so we have 12 ranks left to choose from.
- We need to choose 1 card of one rank and 1 card of another rank, which can be done in (4 choose 1) * (4 choose 1) = 16 ways.
Therefore, the total number of favorable outcomes is 13 * 6 * 16 = 1248.
Total Number of Possible Outcomes:
To calculate the total number of possible outcomes, we need to choose 4 cards from a deck of 52 cards, which can be done in (52 choose 4) = 270,725 ways.
Calculation of Probability:
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
Therefore, the probability of drawing exactly one pair can be calculated as follows:
Probability = Number of Favorable Outcomes / Total Number of Possible Outcomes
= 1248 / 270,725
≈ 0.0046
However, the given correct answer is '0.304', which means the probability needs to be expressed as a percentage. To convert the probability to a percentage, we multiply it by 100.
Probability (as a percentage) = 0.0046 * 100
≈ 0.46%
Therefore, the probability of drawing exactly one pair is approximately 0.46%.
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}
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
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hi t correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is- [AIEEE 2007]- a)0.06
- b)0.14
- c)0.2
- d)0.7
Correct answer is option 'B'. Can you explain this answer?
Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hi t correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane is-
[AIEEE 2007]
a)
0.06
b)
0.14
c)
0.2
d)
0.7
|
Rithika Khanna answered |
Required probability
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?- a)X = 0, 1, 2; yes
- b)X = 2, 1, 3; yes
- c)X = 2, 3, 5; yes
- d)X = 2, 3, 4; no
Correct answer is option 'A'. Can you explain this answer?
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?
a)
X = 0, 1, 2; yes
b)
X = 2, 1, 3; yes
c)
X = 2, 3, 5; yes
d)
X = 2, 3, 4; no
|
Pankaj Sengupta answered |
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls.therefore the possible values of X = 0 , 1 ,2 . Yes , X is a random variable.
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.
An experiment has 10 equally likely outcomes. Let A and B be non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (2008)- a)2, 4 or 8
- b)3, 6 or 9
- c)4 or 8
- d)5 or 10
Correct answer is option 'D'. Can you explain this answer?
An experiment has 10 equally likely outcomes. Let A and B be non-empty events of the experiment. If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is (2008)
a)
2, 4 or 8
b)
3, 6 or 9
c)
4 or 8
d)
5 or 10
|
|
Shreya Dasgupta answered |
We have n (S) = 10, n(A) = 4
Let n(B) = x and n(A ∩ B) = y
Then for A and B to be independent events P(A ∩ B) = P(A) P(B)
Let n(B) = x and n(A ∩ B) = y
Then for A and B to be independent events P(A ∩ B) = P(A) P(B)
⇒ y can be 2 or 4 so that x = 5 or 10
∴ n (B) = 5 or 10
∴ n (B) = 5 or 10
The probability of India winning a test match against west Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test is (1995S)- a)1/8
- b)1/4
- c)1/2
- d)2/3
Correct answer is option 'B'. Can you explain this answer?
The probability of India winning a test match against west Indies is 1/2. Assuming independence from match to match the probability that in a 5 match series India’s second win occurs at third test is (1995S)
a)
1/8
b)
1/4
c)
1/2
d)
2/3
|
Kirti Datta answered |
Wins at least 3 matches can be calculated using the binomial distribution.
The binomial distribution calculates the probability of getting a certain number of successes (India winning a match) in a fixed number of independent Bernoulli trials (5 matches).
To find the probability that India wins at least 3 matches in a 5-match series, we need to calculate the probabilities of winning 3, 4, and 5 matches and then sum them up.
P(India wins at least 3 matches) = P(India wins 3 matches) + P(India wins 4 matches) + P(India wins 5 matches)
P(India wins k matches) = (5Ck) * (1/2)^k * (1/2)^(5-k)
Where (5Ck) is the binomial coefficient that represents the number of ways to choose k successes from 5 trials.
Calculating each term separately:
P(India wins 3 matches) = (5C3) * (1/2)^3 * (1/2)^(5-3) = 10 * (1/8) * (1/8) = 10/64 = 5/32
P(India wins 4 matches) = (5C4) * (1/2)^4 * (1/2)^(5-4) = 5 * (1/16) * (1/2) = 5/32
P(India wins 5 matches) = (5C5) * (1/2)^5 * (1/2)^(5-5) = 1 * (1/32) * (1/1) = 1/32
Now, summing up the probabilities:
P(India wins at least 3 matches) = 5/32 + 5/32 + 1/32 = 11/32
Therefore, the probability that India wins at least 3 matches in a 5-match series against West Indies is 11/32.
The binomial distribution calculates the probability of getting a certain number of successes (India winning a match) in a fixed number of independent Bernoulli trials (5 matches).
To find the probability that India wins at least 3 matches in a 5-match series, we need to calculate the probabilities of winning 3, 4, and 5 matches and then sum them up.
P(India wins at least 3 matches) = P(India wins 3 matches) + P(India wins 4 matches) + P(India wins 5 matches)
P(India wins k matches) = (5Ck) * (1/2)^k * (1/2)^(5-k)
Where (5Ck) is the binomial coefficient that represents the number of ways to choose k successes from 5 trials.
Calculating each term separately:
P(India wins 3 matches) = (5C3) * (1/2)^3 * (1/2)^(5-3) = 10 * (1/8) * (1/8) = 10/64 = 5/32
P(India wins 4 matches) = (5C4) * (1/2)^4 * (1/2)^(5-4) = 5 * (1/16) * (1/2) = 5/32
P(India wins 5 matches) = (5C5) * (1/2)^5 * (1/2)^(5-5) = 1 * (1/32) * (1/1) = 1/32
Now, summing up the probabilities:
P(India wins at least 3 matches) = 5/32 + 5/32 + 1/32 = 11/32
Therefore, the probability that India wins at least 3 matches in a 5-match series against West Indies is 11/32.
Four fair dice D1, D2, D3 and D4 ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously.
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Four fair dice D1, D2, D3 and D4 ; each having six faces numbered 1, 2, 3, 4, 5 and 6 are rolled simultaneously.
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)
The probability that D4 shows a number appearing on one of D1, D2 and D3 is (2012)
a)
b)
c)
d)
|
|
Harsh Singhal answered |
(3*6*5*5+3*6*5+6*1)/6*6*6*6
and 
then P (B∩C) is (2003S)
An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment.If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is[JEE 2008]- a)2, 4 or 8
- b)3, 6 or 9
- c)4 or 8
- d)5 or 10
Correct answer is option 'D'. Can you explain this answer?
An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment.If A consists of 4 outcomes, the number of outcomes that B must have so that A and B are independent, is
[JEE 2008]
a)
2, 4 or 8
b)
3, 6 or 9
c)
4 or 8
d)
5 or 10
|
|
Rajat Patel answered |
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.
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