Hypergeometric Distribution If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers you selected), while the remaining B objects are of the other type (such as lottery numbers you didn’t select), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n - x objects of type B is

\(P\left( x \right) = \frac{{A!}}{{\left( {A - x} \right)!x!}} \times \frac{{B!}}{{\left( {B - n + x} \right)!\left( {n - x} \right)!}} \div \frac{{\left( {A + B} \right)!}}{{\left( {A + B - n} \right)!n!}}\)

In New Jersey’s Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6, B = 43, n = 6, and x = 2.)

In a lottery game, 6 numbers are selected from 1 to 49 and then a winning six-number combination is selected later.

When a sample of small size is selected from a population without replacement, then the hypergeometric distribution is applied to compute the probability.

Suppose a population has A objects of one kind and the remaining B objects of anotherkind.If n objects are selected without replacement, then the probability of getting x objects of A kind and n-x objects of B kind is computed using the given formula:

\(P\left( x \right) = \frac{{A!}}{{\left( {A - x} \right)!x!}} \times \frac{{B!}}{{\left( {B - n + x} \right)!\left( {n - x} \right)!}} \div \frac{{\left( {A + B} \right)!}}{{\left( {A + B - n} \right)!n!}}\)

The total number of numbers available = 49

The 2 kinds of categories considered are A (numbers that got selected) and B(numbers that did not get selected).

Since 6 numbers are selected, A=6 and

\(\begin{aligned}{c}B = n - x\\ = 49 - 6\\ = 43\end{aligned}\)

The number selected (n) is equal to 6.

The numbers are selected without replacement.

Let\(x\)represent selecting numbers that belong to the winning combination of numbers and it is given as\(x = 2\).

The probability of getting exactly two winning numbers is computed below:

\(\begin{aligned}{c}P\left( x \right) = \frac{{A!}}{{\left( {A - x} \right)!x!}} \times \frac{{B!}}{{\left( {B - n + x} \right)!\left( {n - x} \right)!}} \div \frac{{\left( {A + B} \right)!}}{{\left( {A + B - n} \right)!n!}}\\P\left( 2 \right) = \frac{{6!}}{{\left( {6 - 2} \right)!2!}} \times \frac{{43!}}{{\left( {43 - 6 + 2} \right)!\left( {6 - 2} \right)!}} \div \frac{{\left( {6 + 43} \right)!}}{{\left( {6 + 43 - 6} \right)!6!}}\\ = 15 \times 0.008825202\\ = 0.132\end{aligned}\)

Therefore, the probability of getting exactly two winning numbers with one ticket is equal to 0.132.