*42*

The **hypergeometric distribution**Â describes the probability of choosingÂ *k *objects with a certain feature inÂ *n *draws without replacement, from a finite population of sizeÂ *NÂ *that containsÂ *KÂ *objects with that feature.

If aÂ random variableÂ *X*Â follows a hypergeometric distribution, then the probability of choosingÂ *kÂ *objects with a certain feature can be found by the following formula:

**P(X=k) = _{K}C_{k}Â (_{N-K}C_{n-k}) /Â _{N}C_{n}**

where:

**N:Â**population size**K:Â**number of objects in population with a certain feature**n:Â**sample size**k:Â**number of objects in sample with a certain featurenumber of combinations of K things taken k at a time_{K}C_{k}:Â

For example, there are 4 Queens in a standard deck of 52 cards. Suppose we randomly pick a card from a deck, then, without replacement, randomly pick another card from the deck. What is the probability that both cards are Queens?

To answer this, we can use the hypergeometric distribution with the following parameters:

**N:Â**population size = 52 cards**K:Â**number of objects in population with a certain feature = 4 queens**n:Â**sample size = 2 draws**k:Â**number of objects in sample with a certain feature = 2 queens

Plugging these numbers in the formula, we find the probability to be:

**P(X=2) **= _{K}C_{k}Â (_{N-K}C_{n-k}) /Â _{N}C_{n} =Â _{4}C_{2}Â (_{52-4}C_{2-2}) /Â _{52}C_{2} = 6*1/ 1326 =Â **0.00452**.

This should make sense intuitively. If you imagine yourself pulling two cards out of a deck, one after the other, the probability thatÂ *bothÂ *cards are Queens should be very low.

**Properties of the Hypergeometric Distribution**

The hypergeometric distribution has the following properties:

The mean of the distribution is**Â (nK) / N**

The variance of the distribution is **(nK)(N-K)(N-n) / (N ^{2}(n-1))**

**Hypergeometric Distribution Practice Problems**

Use the following practice problems to test your knowledge of the hypergeometric distribution.

**Note:Â **We will use the Hypergeometric Distribution Calculator to calculate the answers to these questions.

**ProblemÂ 1**

**Question:Â **Suppose we randomly pick four cards from a deck without replacement. What is the probability that two of the cards are Queens?

To answer this, we can use the hypergeometric distribution with the following parameters:

**N:Â**population size = 52 cards**K:Â**number of objects in population with a certain feature = 4 queens**n:Â**sample size = 4 draws**k:Â**number of objects in sample with a certain feature = 2 queens

Plugging these numbers into the Hypergeometric Distribution Calculator, we find the probability to beÂ **0.025**.

**ProblemÂ 2**

**Question:Â **An urn contains 3 red balls and 5 green balls. You randomly choose 4 balls. What is the probability that you choose exactly 2 red balls?

To answer this, we can use the hypergeometric distribution with the following parameters:

**N:Â**population size = 8 balls**K:Â**number of objects in population with a certain feature = 3 red balls**n:Â**sample size = 4 draws**k:Â**number of objects in sample with a certain feature = 2 red balls

Plugging these numbers into the Hypergeometric Distribution Calculator, we find the probability to beÂ **0.42857**.

**ProblemÂ 3**

**Question:Â **A basketÂ contains 7 purple marbles and 3 pink marbles. You randomly choose 6 marbles. What is the probability that you choose exactly 3 pink marbles?

To answer this, we can use the hypergeometric distribution with the following parameters:

**N:Â**population size = 10 marbles**K:Â**number of objects in population with a certain feature = 3 pink balls**n:Â**sample size = 6 draws**k:Â**number of objects in sample with a certain feature = 3 pink balls

Plugging these numbers into the Hypergeometric Distribution Calculator, we find the probability to beÂ **0.16667**.