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All questions of Chapter 13 - Probability for Commerce 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 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)
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
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
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
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 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.
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.
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
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.
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.
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is- a)41/133
- b)3/13
- c)1/13
- d)2/13
Correct answer is option 'D'. Can you explain this answer?
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. Then the value of E(X) is
a)
41/133
b)
3/13
c)
1/13
d)
2/13
|
|
Sounak Yadav answered |
Given:
- Two cards are drawn at random from a deck of cards.
- X represents the number of aces obtained.
To find:
- The value of E(X), which is the expected value of X.
Approach:
- We can find the expected value of X by calculating the probability of obtaining each possible outcome and multiplying it by the corresponding value of X.
Calculation:
- Total number of ways to draw 2 cards from a deck of 52 cards: C(52, 2) = 52! / (2!(52-2)!) = 1326
Step 1: Calculate the probability of obtaining 0 aces
- Number of ways to select 2 non-ace cards from the 48 non-ace cards: C(48, 2) = 48! / (2!(48-2)!) = 1128
- Probability of obtaining 0 aces: P(X = 0) = 1128 / 1326
Step 2: Calculate the probability of obtaining 1 ace
- Number of ways to select 1 ace card from the 4 ace cards: C(4, 1) = 4
- Number of ways to select 1 non-ace card from the 48 non-ace cards: C(48, 1) = 48
- Probability of obtaining 1 ace: P(X = 1) = (4 * 48) / 1326
Step 3: Calculate the probability of obtaining 2 aces
- Number of ways to select 2 ace cards from the 4 ace cards: C(4, 2) = 6
- Probability of obtaining 2 aces: P(X = 2) = 6 / 1326
Step 4: Calculate the expected value of X
- Expected value of X (E(X)) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2))
- E(X) = (0 * 1128/1326) + (1 * 4 * 48/1326) + (2 * 6/1326)
- E(X) = 0 + 192/1326 + 12/1326
- E(X) = 204/1326
- E(X) = 2/13
Therefore, the value of E(X) is 2/13. Hence, option D is the correct answer.
- Two cards are drawn at random from a deck of cards.
- X represents the number of aces obtained.
To find:
- The value of E(X), which is the expected value of X.
Approach:
- We can find the expected value of X by calculating the probability of obtaining each possible outcome and multiplying it by the corresponding value of X.
Calculation:
- Total number of ways to draw 2 cards from a deck of 52 cards: C(52, 2) = 52! / (2!(52-2)!) = 1326
Step 1: Calculate the probability of obtaining 0 aces
- Number of ways to select 2 non-ace cards from the 48 non-ace cards: C(48, 2) = 48! / (2!(48-2)!) = 1128
- Probability of obtaining 0 aces: P(X = 0) = 1128 / 1326
Step 2: Calculate the probability of obtaining 1 ace
- Number of ways to select 1 ace card from the 4 ace cards: C(4, 1) = 4
- Number of ways to select 1 non-ace card from the 48 non-ace cards: C(48, 1) = 48
- Probability of obtaining 1 ace: P(X = 1) = (4 * 48) / 1326
Step 3: Calculate the probability of obtaining 2 aces
- Number of ways to select 2 ace cards from the 4 ace cards: C(4, 2) = 6
- Probability of obtaining 2 aces: P(X = 2) = 6 / 1326
Step 4: Calculate the expected value of X
- Expected value of X (E(X)) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2))
- E(X) = (0 * 1128/1326) + (1 * 4 * 48/1326) + (2 * 6/1326)
- E(X) = 0 + 192/1326 + 12/1326
- E(X) = 204/1326
- E(X) = 2/13
Therefore, the value of E(X) is 2/13. Hence, option D is the correct answer.
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of at least 5 successes?- a)7/64
- b)5/64
- c)9/64
- d)3/64
Correct answer is option 'A'. Can you explain this answer?
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of at least 5 successes?
a)
7/64
b)
5/64
c)
9/64
d)
3/64
|
Nabanita Singh answered |
p = probability of success = 3/6 = ½ .
q = probability of failure = 1 – p = 1 – ½ = ½
. let x be the number of successes , then x has the binomial distribution with :
n = 6 , p = ½ , q = ½ .
q = probability of failure = 1 – p = 1 – ½ = ½
. let x be the number of successes , then x has the binomial distribution with :
n = 6 , p = ½ , q = ½ .
There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?- a)4/9
- b)2/9
- c)1/9
- d)5/9
Correct answer is option 'A'. Can you explain this answer?
There are three coins. One is a two headed coin (having head on both faces),another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin ?
a)
4/9
b)
2/9
c)
1/9
d)
5/9
|
Harshitha Nambiar answered |
Let
E1 , E2 and E3 and are events of selection of a two headed coin , biased coin and unbiased coin respectively.
E1 , E2 and E3 and are events of selection of a two headed coin , biased coin and unbiased coin respectively.
Let A = event of getting head. 
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is- a)9/10
- b)(9/10)5
- c)(1/2 )5
- d)10-1
Correct answer is option 'B'. Can you explain this answer?
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
a)
9/10
b)
(9/10)5
c)
(1/2 )5
d)
10-1
|
|
Aarav Pillai answered |
Probability of getting first non-defective bulb =90/100
Probablity of getting second non-defective bulb =89/99
Probablity of getting third non-defective bulb =88/98
Probablity of getting fourth non-defective bulb =87/97
Probablity of getting fifth non-defective bulb =86/96
So, probablity of getting all non-defective bulbs in a sample of 5 bulbs =
(90/100)∗(89/99)∗(88/98)∗(87/97)∗(86/96)
= closest option is b
Probablity of getting second non-defective bulb =89/99
Probablity of getting third non-defective bulb =88/98
Probablity of getting fourth non-defective bulb =87/97
Probablity of getting fifth non-defective bulb =86/96
So, probablity of getting all non-defective bulbs in a sample of 5 bulbs =
(90/100)∗(89/99)∗(88/98)∗(87/97)∗(86/96)
= closest option is b
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?- a)3/52
- b)5/52
- c)1/52
- d)1/4
Correct answer is option 'C'. Can you explain this answer?
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
a)
3/52
b)
5/52
c)
1/52
d)
1/4
|
|
Athira Datta answered |
Let
E1 , E2 and E3 and are events of selection of a scooter driver , car driver and truck driver respectively.
E1 , E2 and E3 and are events of selection of a scooter driver , car driver and truck driver respectively.
In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb is put back in the box before making the second selection. The probability that both the bulbs are not defective is:- a)8/25
- b)4/5
- c)16/25
- d)9/25
Correct answer is option 'C'. Can you explain this answer?
In a box of 10 electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb is put back in the box before making the second selection. The probability that both the bulbs are not defective is:
a)
8/25
b)
4/5
c)
16/25
d)
9/25
|
|
Khaja Moinuddin answered |
If E, F and G are events then P ((E ∪ F)|G) =- a)P (E|G) + P (G|F) – P ((E ∩ F)|G)
- b)P (E|G) + P (F|G) – P ((E ∩ F)|G)
- c)P (E|G) + P (F|G) – P ((E ∩ F)|F)
- d)P (G|E) + P (F|G) – P ((E ∩ F)|E)
Correct answer is option 'B'. Can you explain this answer?
If E, F and G are events then P ((E ∪ F)|G) =
a)
P (E|G) + P (G|F) – P ((E ∩ F)|G)
b)
P (E|G) + P (F|G) – P ((E ∩ F)|G)
c)
P (E|G) + P (F|G) – P ((E ∩ F)|F)
d)
P (G|E) + P (F|G) – P ((E ∩ F)|E)
|
|
Anshika Rane answered |
If E, F and G are events then P ((E ∪ F)|G) represents the conditional probability of the given event . therefore P ((E ∪ F)|G) = P (E|G) + P (F|G) – P ((E ∩ F)|G) .
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?- a)22/133
- b)23/133
- c)27/133
- d)25/133
Correct answer is option 'A'. Can you explain this answer?
A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
a)
22/133
b)
23/133
c)
27/133
d)
25/133
|
Gauri Chauhan answered |
Let
E1 and E2 are events that a person has disease and the person has not disease.
E1 and E2 are events that a person has disease and the person has not disease.
Let A = event that the test result is positive
A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, and scooter or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 1/4 , 1/3 and 1/12, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. The probability that he comes by bus is:- a)1/6
- b)4/9
- c)5/18
- d)1/18
Correct answer is option 'B'. Can you explain this answer?
A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, and scooter or by other means of transport are respectively 0.3, 0.2, 0.1 and 0.4. The probabilities that he will be late are 1/4 , 1/3 and 1/12, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. The probability that he comes by bus is:
a)
1/6
b)
4/9
c)
5/18
d)
1/18
|
Madhurima Rane answered |
The correct answer is B.
Let T1:Event that doctor came by train
T2:Event that doctor came by Bus
T3:Event that doctor came by scooter
T4:Event that doctor came by other means of transport
L :Event that doctor arrives late.
Probability if doctor comes by train if he is late
A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.- a)5/9
- b)1/9
- c)7/9
- d)4/9
Correct answer is option 'B'. Can you explain this answer?
A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
a)
5/9
b)
1/9
c)
7/9
d)
4/9
|
Pranavi Chavan answered |
n(S)=36.
Let A = event of getting sum 8.
= {(2,6),(3,5),(4,4),(5,3),(6,2),}
And B = event of getting a number less than 4 on red die.
={(1,1),(2,1),(3,1),(4,1),(5,1),(5,1), (1,2),(2,2),(3,2),(4,2),(5,2),(6,2), (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)}.
Let A = event of getting sum 8.
= {(2,6),(3,5),(4,4),(5,3),(6,2),}
And B = event of getting a number less than 4 on red die.
={(1,1),(2,1),(3,1),(4,1),(5,1),(5,1), (1,2),(2,2),(3,2),(4,2),(5,2),(6,2), (1,3),(2,3),(3,3),(4,3),(5,3),(6,3)}.
Direction: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice asAssertion (A) : The probability of an impossible event is 1.Reason (R) : If A is a perfect subset of B and P(A) < p(b),="" then="" p(b="" –="" a)="" is="" equal="" to="" p(b)="" –="" />- a)Both A and R are true and R is the correct explanation of A
- b)Both A and R are true but R is NOT the correct explanation of A
- c)A is true but R is false
- d)A is false but R is True
Correct answer is option 'D'. Can you explain this answer?
Direction: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as
Assertion (A) : The probability of an impossible event is 1.
Reason (R) : If A is a perfect subset of B and P(A) < p(b),="" then="" p(b="" –="" a)="" is="" equal="" to="" p(b)="" –="" />
a)
Both A and R are true and R is the correct explanation of A
b)
Both A and R are true but R is NOT the correct explanation of A
c)
A is true but R is false
d)
A is false but R is True
|
|
Anjali Sharma answered |
Assertion (A) is wrong.
If the probability of an event is 0, then it is called an impossible event.
But Reason (R) is correct.
From Basic Theorem of Probability, P(B – A) = P(B) – P(A), this is true only if the condition given in the question is true.
The conditional probability of an event E’s complement E’, given the occurrence of the event F- a)P (E′|F) = 1
- b)P (E′|F) = 1 – P (E|F)
- c)P (E′|F) = – 1 + P (E|F)
- d)P (E′|F) = P (E|F)
Correct answer is option 'B'. Can you explain this answer?
The conditional probability of an event E’s complement E’, given the occurrence of the event F
a)
P (E′|F) = 1
b)
P (E′|F) = 1 – P (E|F)
c)
P (E′|F) = – 1 + P (E|F)
d)
P (E′|F) = P (E|F)
|
Anuj Saini answered |
As the total probability of an event is always 1 . therefore , P (E′|F) = 1 – P (E|F).
Which of the following conditions do Bernoulli trials satisfy?- a)finite number of dependent trials
- b)infinite number of dependent trials.
- c)infinite number of independent trials
- d)finite number of independent trials
Correct answer is option 'D'. Can you explain this answer?
Which of the following conditions do Bernoulli trials satisfy?
a)
finite number of dependent trials
b)
infinite number of dependent trials.
c)
infinite number of independent trials
d)
finite number of independent trials
|
|
Anjali Deshpande answered |
Bernoulli trials satisfies the finite number of independent trials .
A bag ‘A’ contains 2 white and 3 red balls and a bag ‘B’ contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. The probability that it was drawn from bag B is:- a)27/52
- b)3/5
- c)25/52
- d)5/9
Correct answer is option 'C'. Can you explain this answer?
A bag ‘A’ contains 2 white and 3 red balls and a bag ‘B’ contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. The probability that it was drawn from bag B is:
a)
27/52
b)
3/5
c)
25/52
d)
5/9
|
Swati Khanna answered |
If we assume that it is equally likely that the ball is drawn from either bag, then we proceed as follows. Let R be the event that a red ball is drawn, and let B be the event that bag B is chosen for the drawing. Then
P(B) = P(not B) = 1/2
P(R|B) = 5/9 = P(RB) / P(B) = 2 P(RB),
P(R|not B) = 3/5 = P(R not B) / P(not B) = 2 P(R not B)
P(R) = P(RB) + P(R not B) = (1/2)( 5/9 + 3/5 ) = 26/45
P(B|R) = P(RB) / P(R) = (5/9)(1/2) / (26/45) = 25/52
P(B) = P(not B) = 1/2
P(R|B) = 5/9 = P(RB) / P(B) = 2 P(RB),
P(R|not B) = 3/5 = P(R not B) / P(not B) = 2 P(R not B)
P(R) = P(RB) + P(R not B) = (1/2)( 5/9 + 3/5 ) = 26/45
P(B|R) = P(RB) / P(R) = (5/9)(1/2) / (26/45) = 25/52
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. The probability that there is at least one defective egg is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. The probability that there is at least one defective egg is:
a)
b)
c)
d)
|
Geetika Shah answered |
Probability of an egg being defective =10/100=110
So, probability of an egg being non-defective=1−0.1=0.9
10 eggs are drawn successively with replacement.
So, the probability of getting no defective egg =(0.9)10
Hence, the probability that there is at least one defective egg = 1−(0.9)10
So, probability of an egg being non-defective=1−0.1=0.9
10 eggs are drawn successively with replacement.
So, the probability of getting no defective egg =(0.9)10
Hence, the probability that there is at least one defective egg = 1−(0.9)10
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