Hypergeometric Distribution [Probability]

Hi there. In this math post, I cover the hypergeometric distribution from probability.

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Topics

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• Review Of Combinations
• Hypergeometric Distribution
• Examples

Review Of Combinations

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In the math version of combinations, it is referred to how many ways k items can be selected out of a larger pool of n choices. With math notation, it is defined as:

 

Note that the factorial k! is defined as: for k 1 or greater. Zero factorial 0! is defined to be equal to just 1.

Example One

There are 5 pizza slice options at the local pizza place. John wants 2 different pizza slices from the 5. How many ways can John select 2 different slices from the 5 choices?

There are 10 ways to choose 2 different slices from 5 choices. On a calculator, use 5C2. You may need the second function or SHIFT part on the calculator to access the choose function.

 

Example Two

Billy wants to have 4 teammates for a quick game of Counter Strike against another 5 man team. He has 7 people to choose from. How many different ways can Billy choose 4 from 7 candidates? Note that order does not matter here.

 

There are 35 ways of choosing 4 teammates from 7 candidates. If you have a calculator you can do 7C4.

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Hypergeometric Distribution

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Consider two groups where the first group has X many choices and the second group has Y many choices. You want to choose x items or people from X many choices in the first group. Also, you want to choose yitems/people from the second Y group.

The probability of choosing x items from X choices and choosing y items from Y choices is as follows:

 

The denominator is the amount of ways you can choose x+y from X + Y options. Note that X + Y many options represent all the choices available.

This formula will make more sense through some examples.

 

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Examples

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Example One

There are 5 men and 3 women as candidates. What is the probability of choosing 3 men and 1 woman in a 4 person committee?

 

The number of ways of choosing 3 men from 5 is 5 choose 3 which is equal to 10 ways.

Choosing one woman from 3 women is 3 choose 1 or 3 ways.

Multiplying 10 and 3 gives the numerator of 30.

The denominator is the 8 candidates choosing 4 people for the committee. This is 8 choose 4 which is equal to 70.

The probability of choosing 3 men and 1 woman for the 4 person committee is 30/70 or 3/7 which is approximately 42.86%.

Here are the calculations:

 

Example Two

You have a standard deck of 52 cards. There are 13 cards for each suit. You draw 5 cards from the 52 card without replacement. What is the probability of all 5 cards being hearts (flush)?

Start with the denominator. It is 52 cards choose 5. Computing 52 choose 5 on the calculator gives 2,598,960 ways of choosing 5 cards from 52 cards.

For the numerator there are two things to consider. For the hearts, consider selecting 5 hearts out of a possible 13. This is 13 choose 5 which is 1287 ways. As the 5 cards are all hearts, you have 39 cards that are not hearts and you choose none of these 39 in the selected 5 cards. There is only way to choose no cards from 39.

Altogether the probability of all 5 selected cards being hearts is as follows:

 

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Thank you for reading.

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