Question

A club is choosing 2 members to serve on a committee. The club has nominated 2 women and 4 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

Answer:

53.33% probability that one woman and one man will be chosen to be on the committee

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The order in which the members are chosen is not important, so we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

What is the probability that one woman and one man will be chosen to be on the committee?

Desired outcomes:

One woman, from a set of 2, and one man, from a set of 4. So

$D = C_{2,1}*C_{4,1} = \frac{2!}{1!1!}*\frac{4!}{1!3!} = 8$

Total outcomes:

Two members from a set of 2 + 4 = 6. So

$T = C_{6,2} = \frac{6!}{2!4!} = 15$

Probability:

$p = \frac{D}{T} = \frac{8}{15} = 0.5333$

53.33% probability that one woman and one man will be chosen to be on the committee