Probability and Statistics Multiple Choice Questions & Answers (MCQs) 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
Answer: d
Clarification: 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
Answer: a
Clarification: 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
Answer: a
Clarification: 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) = [kCx] [N-kCn-x] / [NCn]
h(3; 52, 6, 26) = [26C3] [26C3] / [52C6]
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
Answer: c
Clarification: 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) = [kCx] [N-kCn-x] / [NCn]
where x ranges from 0 to 2.
h(x<3; 52, 8, 13) = [13C0] [39C8] / [52C8] + [13C1] [39C7] / [52C8] + [13C2] [39C6] / [52C2]
h(x<3; 52, 8, 13) = 0.685.
5. The Variance of hypergeometric distribution is given as __________
a) n * k * (N – k) * (N – 1) / [N2 * (N – 1)]
b) n * k * (N – k) * (N – n) / [N2 * (N – k)]
c) n * k * (N – 1) * (N – n) / [N2 * (N – 1)]
d) n * k * (N – k) * (N – n) / [N2 * (N – 1)]
Answer: d
Clarification: The variance of hypergeometric distribution is given as n * k * (N – k) * (N – n) / [N2 * (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) = [kCx] [NCn-x] / [NCn]
b) h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
c) h(x; N, n, k) = [kCx] [N-kCn] / [NCn]
d) h(x; N, n, k) = [kCx] [N-kCn-x] / [N-kCn]
Answer: b
Clarification: Hypergeometric probability of hypergeometric distribution function is given by the formula:
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
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
Answer: d
Clarification: The Variance of hypergeometric distribution is given as,
n * k * (N – k) * (N – 1) / [N2 * (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
Answer: d
Clarification: 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) = [kCx] [N-kCn-x] / [NCn]
h(3; 52, 4, 4) = [4C3] [48C1] / [52C4]
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
Answer: a
Clarification: 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
Answer: b
Clarification: 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) = [kCx] [N-kCn-x] / [NCn]
h(0; 52, 4, 3) = [4C0] [48C3] / [52C3]
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
Answer: c
Clarification: The Variance of hypergeometric distribution is given as,
n * k * ( N – k ) * ( N – 1 ) / [N2 * (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
Answer: b
Clarification: 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
Answer: a
Clarification: 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) = [kCx] [N-kCn-x] / [NCn]
h(2; 52, 5, 36) = [36C2] [12C3] / [52C5]
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
Answer: a
Clarification: 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.