91Ó°ÊÓ

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.