Probability Test

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Probability Test - 10

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Question 1
1 / -0.25

The variance of the distribution is

A
21/100

B
30/100

C
49/100

D
35/100

Solution

Mean= E(X) =
= 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
Question 2
1 / -0.25

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

Solution

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.
Question 3
1 / -0.25

A man make attempts to hit the target. The probability of hitting the target is 3/5 . Then the probability that A hit the target exactly 2 times in 5 attempts is:

A
72/3125

B
216/625

C
144/3125

D
144/625

Solution

Probability of success(p) = ⅗
q = 1 - p
= 1 - (⅗) ⇒⅖
Probability of 2 hit in 5 attempts = 5C2 (⅗)² (⅖)³
= 144/625
Question 4
1 / -0.25

Trials of a random experiment are called Bernoulli trials, if they satisfy the condition/s:

A
There should be a finite number of independent trials

B
Each trial has exactly two outcomes: success or failure.

C
The probability of success remains the same in each trial.

D
All of the above
Question 5
1 / -0.25

In binomial probability distribution, mean is 3 and standard deviation is 3/2 . Then the probability distribution is:

A

B

C

D

Solution

Mean = np = 3, S.D. (npq)^½ = 3/2

p = ¼ n = 12
Hence, binomial distribution is (p+q)ⁿ = (¾+ ¼)¹²
Question 6
1 / -0.25

A random variable is a real valued function whose domain is the.

A
set of integers

B
set of irrational numbers

C
sample space of a random experiment

D
set of real numbers
Question 7
1 / -0.25

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

Solution

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
Question 8
1 / -0.25

The mean and variance of a binomial distribution are 4 and 3 respectively, then the probability of getting exactly six successes in this distribution is:

A

B

C

D

Solution

Let X ˜B(n,p) be a binomial variate with mean 4 and variance 3. Then,
np=4,and npq=3
⇒q=3/4,p=1/4 and n=16
∴P(X=r)=.16Cr(1/4)^r (3/4)^{(16 −r)} ,r=0,1,2,....,16
⇒P(X=6)= 16C6(1/4)⁶ ((3)/(4))¹⁰
Question 9
1 / -0.25

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
Question 10
1 / -0.25

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

Solution

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