Question

A club is choosing 2 members to serve on a committee. The club has nominated 2 women and 2 men. Based on chance alone, what is the probability that one woman and one man will be chosen to be on the committee?

Answer 1

Probability is used to determine the chances of an event. The probability of that one man and one woman will be in the committee is 0.667

Given

$Women = 2$

$Men = 2$

$Total = 2 + 2 = 4$

The number of ways of selecting 2 members from the 4 nominated people is:

$^nC_r = \frac{n!}{(n - r)!r!}$

Where:

$n = 4$ --- the committee members

$r = 2$ --- the members to be selected

So, we have:

$^4C_2 = \frac{4!}{(4 - 2)!2!}$

$^4C_2 = \frac{4!}{2!2!}$

$^4C_2 = \frac{4 \times 3 \times 2!}{2!2 \times 1}$

$^4C_2 = \frac{4 \times 3}{2}$

$^4C_2 = \frac{12}{2}$

$^4C_2 = 6$

The number of ways of selecting 1 man from 2 is:

$^2C_1 = \frac{2!}{(2 - 1)!1!} = 2$

The number of ways of selecting 1 woman from 2 is:

$^2C_1 = \frac{2!}{(2 - 1)!1!} = 2$

So, the probability of selecting one man and one woman is:

$Pr = \frac{2 \times 2}{6}$

$Pr = \frac{4}{6}$

$Pr = 0.667$

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