ES1001 : Hypergeometric Distributions

The mean of hypergeometric distribution is _____________

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.

Answer explanation

Explanation:            

E(X) = n*k /N

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

Hence E (X) = 1/2

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

• h(x; N, n, k) =

  [^kC_x] [^NC_n-x] / [^NC_n]

• h(x; N, n, k) =

  [^kC_x] [^N-kC_n-x] / [^NC_n]

• h(x; N, n, k) =

  [^kC_x] [^N-kC_n] / [^NC_n]

• h(x; N, n, k) =

  [^kC_x] [^N-kC_n-x] / [^N-kC_n]

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

• 0.3320

• 0.3240

• 0.4320

• 0.5430

Answer explanation

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) = [^kC_x] [^N-kC_n-x] / [^NC_n]
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

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

• 0.985

• 0.785

• 0.685

• 0.585

Answer explanation

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) = [^kC_x] [^N-kC_n-x] / [^NC_n]
h(x<3; 52, 8, 13) = [¹³C₀] [³⁹C₈] / [⁵²C₈] + [¹³C₁] [³⁹C₇] / [⁵²C₈] + [¹³C₂] [³⁹C₆] / [⁵²C₂]
h(x<3; 52, 8, 13) = 0.685

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

• 0.9999

• 0.9997

• 0.0009

• 0.0007

Answer explanation

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) = [^kC_x] [^N-kC_n-x] / [^NC_n]
h(3; 52, 4, 4) = [⁴C₃] [⁴⁸C₁] / [⁵²C₄]
h(3; 52, 4, 4) = 0.0007

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

• 0.8762

• 0.7826

• 0.8726

• 0.7862

Answer explanation

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) = [^kC_x] [^N-kC_n-x] / [^NC_n]

h(0; 52, 4, 3) = [⁴C₀] [⁴⁸C₃] / [⁵²C₃]

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

1.       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?

• 0.0533

• 0.0753

• 0.0633

• 0.6573

Answer explanation

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) = [^kC_x] [^N-kC_n-x] / [^NC_n]

h(2; 52, 5, 36) = [³⁶C₂] [¹²C₃] / [⁵²C₅]

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