This set of Probability and Statistics Multiple Choice Questions & Answers (MCQs) focuses on “Hypergeometric Distributions”.

1. The mean of hypergeometric distribution is _____________

a) n*k /N-1

b) n*k-1 /N

c) n-1*k /N

d) n*k /N

View Answer

Explanation: The mean of hypergeometric distribution is given as

E(X) = n*k /N where,

n is the number of trials, k is the number of success and N is the sample size.

2. The probability of success and failures in hypergeometric distribution is not fixed.

a) True

b) False

View Answer

Explanation: In Binomial Distribution the probability of success and failures has to be fixed.

On the other hand hypergeometric distribution probability of success and failures is not fixed.

3. Consider selecting 6 cards from a pack of cards without replacement. What is the probability that 3 of the cards will be black?

a) 0.3320

b) 0.3240

c) 0.4320

d) 0.5430

View Answer

Explanation: The given Experiment follows Hypergeometric distribution with

N = 52 since there are 52 cards in a deck.

k = 26 since there are 26 black cards in a deck.

n = 6 since we randomly select 6 cards from the deck.

x = 3 since 3 of the cards we select are black.

h(x; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}]

h(3; 52, 6, 26) = [

^{26}C

_{3}] [

^{26}C

_{3}] / [

^{52}C

_{6}]

h(3; 52, 6, 26) = 0.3320

Thus, the probability of randomly selecting 6 black cards is 0.3320.

4. Suppose we draw eight cards from a pack of 52 cards. What is the probability of getting less than three spades?

a) 0.985

b) 0.785

c) 0.685

d) 0.585

View Answer

Explanation: The Random Experiment follows hypergeometric distribution with,

N = 52 since there are 52 cards in a deck.

k = 13 since there are 13 spades in a deck.

n = 8 since we randomly select 8 cards from the deck.

x = 0 to 2 since we want less than 3 spades.

h(x<3; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}] where x ranges from 0 to 2.

h(x<3; 52, 8, 13) = [

^{13}C

_{0}] [

^{39}C

_{8}] / [

^{52}C

_{8}] + [

^{13}C

_{1}] [

^{39}C

_{7}] / [

^{52}C

_{8}] + [

^{13}C

_{2}] [

^{39}C

_{6}] / [

^{52}C

_{2}]

h(x<3; 52, 8, 13) = 0.685.

5. The Variance of hypergeometric distribution is given as __________

a) n * k * (N – k) * (N – 1) / [N^{2} * (N – 1)]

b) n * k * (N – k) * (N – n) / [N^{2} * (N – k)]

c) n * k * (N – 1) * (N – n) / [N^{2} * (N – 1)]

d) n * k * (N – k) * (N – n) / [N^{2} * (N – 1)]

View Answer

Explanation: The variance of hypergeometric distribution is given as n * k * (N – k) * (N – n) / [N

^{2}* (N – 1)] where,

n is the number of trials, k is the number of success and N is the sample size.

6. Hypergeometric probability of hypergeometric distribution function is given by the formula _________

a) h(x; N, n, k) = [^{k}C_{x}] [^{N}C_{n-x}] / [^{N}C_{n}]

b) h(x; N, n, k) = [^{k}C_{x}] [^{N-k}C_{n-x}] / [^{N}C_{n}]

c) h(x; N, n, k) = [^{k}C_{x}] [^{N-k}C_{n}] / [^{N}C_{n}]

d) h(x; N, n, k) = [^{k}C_{x}] [^{N-k}C_{n-x}] / [^{N-k}C_{n}]

View Answer

Explanation: Hypergeometric probability of hypergeometric distribution function is given by the formula:

h(x; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}]

Where n is the number of trials, k is the number of success and N is the sample size.

7. Find the Variance of a Hypergeometric Distribution such that the probability that a 3-trial hypergeometric experiment results in exactly 2 successes, when the population consists of 7 items.

a) 0.6212

b) 0.6612

c) 0.6112

d) 0.6122

View Answer

Explanation: The Variance of hypergeometric distribution is given as,

n * k * (N – k) * (N – 1) / [N

^{2}* (N – 1)] where,

n is the number of trials, k is the number of success and N is the sample size.

Hence n = 3, k = 2, N = 7.

Var(X) = 0.6122.

8. Suppose we draw 4 cards from a pack of 52 cards. What is the probability of getting exactly 3 aces?

a) 0.9999

b) 0.9997

c) 0.0009

d) 0.0007

View Answer

Explanation: The Random Experiment follows hypergeometric distribution with,

N = 52 since there are 52 cards in a deck.

k = 4 since there are 4 aces in a deck.

n = 4 since we randomly select 4 cards from the deck.

x = 3 since we want 3 aces.

h(x; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}]

h(3; 52, 4, 4) = [

^{4}C

_{3}] [

^{48}C

_{1}] / [

^{52}C

_{4}]

h(3; 52, 4, 4) = 0.0007.

9. The trials conducted in Hypergeometric distribution are done without replacement of the drawn samples.

a) True

b) False

View Answer

Explanation: The trials conducted in Hypergeometric distribution are done without replacement of the drawn samples hence the probability of success and failure is not fixed.

10. Consider Nick draws 3 cards from a pack of 52 cards. What is the probability of getting no kings?

a) 0.8762

b) 0.7826

c) 0.8726

d) 0.7862

View Answer

Explanation: The Random Experiment follows hypergeometric distribution with,

N = 52 since there are 52 cards in a deck.

k = 4 since there are 4 kings in a deck.

n = 3 since we randomly select 3 cards from the deck.

x = 0 since we want no kings.

h(x; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}]

h(0; 52, 4, 3) = [

^{4}C

_{0}] [

^{48}C

_{3}] / [

^{52}C

_{3}]

h(0; 52, 4, 3) = 0.7826.

11. Find the Variance of a Hypergeometric Distribution such that the probability that a 6-trial hypergeometric experiment results in exactly 4 successes, when the population consists of 10 items.

a) 14.4

b) 144

c) 1.44

d) 0.144

View Answer

Explanation: The Variance of hypergeometric distribution is given as,

n * k * ( N – k ) * ( N – 1 ) / [N

^{2}* (N – 1)] where,

n is the number of trials, k is the number of success and N is the sample size.

Hence n = 6, k = 4, N = 10.

Var(X) = 1.44.

12. Hypergeometric Distribution is Continuous Probability Distribution.

a) True

b) False

View Answer

Explanation: Hypergeometric Distribution is a Discrete Probability Distribution. It defines the probability of k successes in n trials from N samples.

13. Emma likes to play cards. She draws 5 cards from a pack of 52 cards. What is the probability of that from the 5 cards drawn Emma draws only 2 face cards?

a) 0.0533

b) 0.0753

c) 0.0633

d) 0.6573

View Answer

Explanation: The Random Experiment follows hypergeometric distribution with,

N = 52 since there are 52 cards in a deck.

k = 36 since there are 36 face cards in a deck.

n = 5 since we randomly select 5 cards from the deck.

x = 2 since we want 2 face cards.

h(x; N, n, k) = [

^{k}C

_{x}] [

^{N-k}C

_{n-x}] / [

^{N}C

_{n}]

h(2; 52, 5, 36) = [

^{36}C

_{2}] [

^{12}C

_{3}] / [

^{52}C

_{5}]

h(2; 52, 5, 36) = 0.0533.

14. Find the Expectation of a Hypergeometric Distribution such that the probability that a 4-trial hypergeometric experiment results in exactly 2 successes, when the population consists of 16 items.

a) 1/2

b) 1/4

c) 1/8

d) 1/3

View Answer

Explanation: In Hypergeometric Distribution the Mean or Expectation E(X) is given as

E(X) = n*k /N

Here n = 4, k = 2, N = 16.

Hence E (X) = 1/2.

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