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

Answer:

147/572

Step-by-step explanation:

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

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

Choose 4 women from the 7:

$\binom{7}{4}=\frac{7!}{4!(7-4)!}=35$  ways

Choose 3 men from 9:

$\binom{9}{3}=\frac{9!}{3!(9-3)!}=84$  ways

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

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

Probability:

$\frac{2940}{11440}=\frac{147}{572}$

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

$\binom{9}{3}=\frac{9!}{3!(9-3)!}=\frac{9!}{3!6!}$

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

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