Question

There are 7 women and 9 men with a chance to be on a game show. The producer of the show is going to choose 7 of these people at random to be contestants. What is the probability that the producer chooses 4 women and 3 men? Round your answer to three decimal places.

Answer 1

147/572

Explanation:

The number of combinations of n things chosen r at a time is

`\binom{n}{r}=(n!)/(r!(n-r)!)` (Some writers use the symbol
`_nC_r`)

Choose 4 women from the 7:

`\binom{7}{4}=(7!)/(4!(7-4)!)=35` ways

Choose 3 men from 9:

`\binom{9}{3}=(9!)/(3!(9-3)!)=84` ways

There are 35 x 84 = 2940 ways to choose both, out of

`\binom{16}{7}=(16!)/(7!(16-7)!)=11440` ways to choose any 7 people from the group of 16.

Probability:

`(2940)/(11440)=(147)/(572)`

P.S. As an example of how to calculate combinations, here's the calculation (by hand, a calculator is easier!) of
`\binom{9}{3}`.

`\binom{9}{3}=(9!)/(3!(9-3)!)=(9!)/(3!6!)`

When you write out the factorials, you can do a bunch of cancellation between the numerator and denominator.

`\frac{9\cdot8\cdot7\cdot\cancel{6\cdot5\cdot4\cdot3\cdot2\cdot1}}{(3\cdot2\cdot1)(\cancel{6\cdot5\cdot4\cdot3\cdot2\cdot1)}}=(9\cdot8\cdot7)/(3\cdot2\cdot1)=84`