The hypergeometric probability distribution is used in situations where items are selected and NOT replaced.

In what follows, we will use the mathematical formula for combinations
given by
\[ \displaystyle {M \choose m} = \dfrac{M!}{m!(M-m)!} \quad , \quad \text{read as } \quad \text{"} m \quad\text{choose} \quad M \text{"} \].

where \( M! = 1 \times 2 \times 3 ... \times (M-1) \times M \) , read as "M factorial".

The above formula gives the number of ways \( m \) items are selected from \( M \) items without repetitions. ( \( m \le M \) )

The hypergeometric formula is better explained through a question.

Question

A box contains \( N \) balls of which \( R \) are red balls and the remaining ones are blue balls.

\( n \) balls are selected (without replacement) from the box at random. What is the probability that \( x \) balls from the \( n \) balls selected are red?

__Solution__

There are \( \displaystyle {N \choose n} \) ways of selecting \( n \) balls from a total of \( N \) balls

There are \( \displaystyle {R \choose x} \) ways of selecting \( x \) red balls from a total of \( R \) red balls

If \( N \) is the total number of balls and \( R \) are red, then \( N - R \) are blue.

If \( x \) balls out of the \( n \) are red, then \( n - x \) are blue.

Hence there are \( \displaystyle {N - R \choose n - x} \) ways of selecting \( n - x \) blue balls from a total of \( N - R \) blue balls.

The diagram below explains visually the situation described above.

The number of ways of selecting \( x \) red balls from a total of \( R \) red balls and selecting \( n - x \) blue balls from a total of \( N - R \) blue balls is given by the counting principle as the product

\( \displaystyle {R \choose x} \displaystyle {N - R \choose n - x} \)

Hence using the classical probability formula, the probability that \( x \) balls from the \( n \) balls selected are red is given by

\[ P(X = x) = \dfrac{ \displaystyle {R \choose x} \displaystyle {N - R \choose n - x} }{ \displaystyle {N \choose n} } \]

Example 1

Four balls are to be randomly selected from a box containing 5 red balls and 3 white balls.

a) What is the probability that an equal number of red and white balls are selected?

b) What is the probability that at least 2 red balls are selected?

__Solution to Example 1__

a)

There is a total of 8 balls; hence \( N = 8 \).

There are 5 red balls, hence \( R = 5 \) and \( N - R = 3 \) white balls.

An equal number of red and white balls when \( n = 4 \) are randomly selected means \( x = 2 \) __red__ and \( n - x = 2 \) __white__.

The number of ways of selecting 2 red from a total of 5 red is given by \( { R \choose x} = { 5 \choose 2} \)

The number of ways of selecting 2 white from a total of 3 white is given by \( {N-R \choose x - 2} = {3 \choose 2} \)

The number of ways of selecting 4 balls from a total of 8 balls is given by \( {N \choose n} = {8 \choose 4} \)

Let \( X \) be the number of red balls selected.

Hence

\( P(X = 2) = \dfrac{{5 \choose 2}{3 \choose 2} }{{8 \choose 4}} = 3/7 \)

b)

\( P (X \ge 2) = P (X = 2 ) + P (X = 3 ) + P (X = 4 ) \)

\( = \dfrac{{5 \choose 2}{3 \choose 2} }{{8 \choose 4}} + \dfrac{{5 \choose 3}{3 \choose 1} }{{8 \choose 4}} + \dfrac{{5 \choose 4}{3 \choose 0} }{{8 \choose 4}} \)

\( = 3/7 + 3/7 + 1/14 = 13/14 \)

Example 2

A quality control inspector is to inspect 15 tools 2 of which are defective by selecting 4 at random.

a) What is the probability that there will be one defective tool among the 4 randomly selected?

b) What is the probability that there will be no defective tools among the 4 randomly selected?

c) How many non defective tools do we need to add, to the tools to inspect, in order that the probability of having non defective tools among the 4 selected is 0.70?

__Solution to Example 2__

a)

There are 15 tools with 2 defective and 13 non defective.

The probability that there will be one defective tool among the 4 randomly means \( x = 3 \) are non defective. Hence

\( P(X = 3) = \dfrac{{13 \choose 3}{2 \choose 1} }{{15 \choose 4}} = 0.41905 \)

b)

All four tools are non defective. Hence

\( P(X = 4) = \dfrac{{13 \choose 4}{2 \choose 0} }{{15 \choose 4}} = 0.52381 \)

b)

Let y be the number of non defective tools to be added so that \( P(X = 4) = 0.70\) . Hence

\( P(X = 4) = \dfrac{{13 + y \choose 4}{2 \choose 0} }{{15 + y \choose 4}} = 0.70 \)

We now need to solve the equation

\(\dfrac{{13 + y \choose 4}{2 \choose 0} }{{15 + y \choose 4}} = 0.70 \)

Use formula for combinations

\( \dfrac{\dfrac{(13 + y )!}{4!(13+y-4)!}} {\dfrac{(15 + y )!}{4!(15+y-4)!}} = 0.70 \)

Simplify

\( {\dfrac{(13 + y )!}{(9+y)!}} {\dfrac{(11+y)!}{(15 + y )!}} = 0.70 \)

Note that \( (15 + y )! = (13 + y )! (14+y)(15+y ) \)

and \( (11 + y )! = (9 + y )! (10+y)(11+y ) \)

and simplify

\( \dfrac{(10+y)(11+y )}{ (14+y)(15+y )} = 0.70 \)

Cross multiply

\( (10+y)(11+y ) = 0.7 (14+y)(15+y ) \)

Expand and group like terms and rewrite the equation in standard form

\( 0.3y^2+0.7y-37 = 0 \)

Solve for \( y \) and select positive solution

\( y = 10 \)

We need to add 10 non defective tools for the probability to reach 0.7.

Example 3

A committee of 6 people is to be selected at random from a group of 4 men and 8 women.

a) What is the probability that an equal number of men and women will be in the committee?

a) What is the probability that at most 3 men will be in the committee?

__Solution to Example 3__

a)

There is a total of 12 people (4 men and 8 women); hence \( N = 12 \).

Equal number of men and women \( n = 6 \) are randomly selected means \( x = 3 \) men and \( n - x = 3 \) women.

Let \( X \) be the number of men selected.

Hence

\( P(X = 3) = \dfrac{{4 \choose 3}{8 \choose 3} }{{12 \choose 6}} = 8/33 \)

b)

\( P (X \le 3) = P (X = 0 ) + P (X = 1 ) + P (X = 2 ) + P (X = 3 ) \)

\( = \dfrac{{4 \choose 0}{8 \choose 6} }{{12 \choose 6}} + \dfrac{{4 \choose 1}{8 \choose 5} }{{12 \choose 6}} + \dfrac{{4 \choose 2}{8 \choose 4} }{{12 \choose 6}} + \dfrac{{4 \choose 3}{8 \choose 3} }{{12 \choose 6}}\)

\( = 1/33 + 8/33 + 5/11 + 8/33 = 32/33 \)

Example 4

Six balls are to be randomly selected from a box containing 6 red balls, 4 blue balls and 5 white balls.

a) What is the probability that 3 red, 2 blue and 1 white balls are selected?

__Solution to Example 4__

a)

There is a total of 15 balls; hence \( N = 15 \) and we select \( 6 \); hence

There are \( {15 \choose 6} \) ways to select 6 balls out of 15

There are \( {6 \choose 3} \) ways to select 3 red out of 6

There are \( {4 \choose 2} \) ways to select 2 blue out of 4

There are \( {5 \choose 1} \) ways to select 1 white out of 5

Let \( X \) be the number of red balls selected.

Hence

\( P(X = 3) = \dfrac{{6 \choose 3}{4 \choose 2} {5 \choose 1}}{{15 \choose 6}} = 120/1001 \)

Example 5

Seven numbers are to be randomly selected from the integers between 1 and 49 inclusive. Let \( P(X = x) \) be the probability of having \( x \) even numbers among the 7 selected.

For what value of \( x \) is \( P(X=x) \) the highest?

__Solution to Example 5__

a)

There are 24 even numbers between 1 and 49 inclusive and they are: 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48

There are 25 odd numbers between 1 and 49 inclusive and they are: 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49

There are \( {49 \choose 7} \) ways to select 7 numbers of 49

If \( x \) out of 7 are even, then \( 7 - x \) are odd.

There are \( {24 \choose x} \) ways to select \( x \) even numbers from the 24 odd listed above.

There are \( {25 \choose 7-x} \) ways to select \( 7-x \) odd numbers from the 25 odd listed above.

The probability \( P(X = x) \) is given by

\( P(X = x) = \dfrac{{24 \choose x}{25 \choose 7 - x}}{{49 \choose 7}} \)

We now calculate \( P(X = x) \) for all possible values of \( x \).

\( P(X = 0) = \dfrac{{24 \choose 0}{25 \choose 7}}{{49 \choose 7}} = 0.00560 \)

\( P(X = 1) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.04948 \)

\( P(X = 2) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.170708 \)

\( P(X = 3) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.29806 \)

\( P(X = 4) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.284513 \)

\( P(X = 5) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.148441 \)

\( P(X = 6) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.039172 \)

\( P(X = 7) = \dfrac{{24 \choose 3}{25 \choose 4}}{{49 \choose 7}} = 0.004029 \)

The probability distribution is graphed below and the probability of having \( x = 3 \) even numbers among the 7 selected is the highest.

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