Question

OLA group consists of six men and seven women. Four people are selected to attend a conference.a. In how many ways can four people be selected from this group of thirteen?b. In how many ways can four women be selected from the seven women?c. Find the probability that the selected group will consist of all women.a. The number of ways to select four people from the group of thirteen isb. The number of ways to select four women from the group of seven women isc. The probability is(Type an integer or a simplified fraction.)

Answer 1

Hello!

This is an exercise about combinations. To solve it, we must remember the formula below:

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

• n,: total of elements

,

• p,: number of elements of the group that we need

Knowing it, let's solve your exercise:

a. In how many ways can four people be selected from this group of thirteen?

$\begin{gathered} C_{13,4}=\frac{13!}{4!(13-4)!}=\frac{13!}{4\times3\times2\times1\times(9!)} \\ \\ =\frac{13\times12\times11\times10\times\cancel{9!}}{4\times3\times2\times1\times(\cancel{9!})}=\frac{17160}{24}=715 \end{gathered}$

b. In how many ways can four women be selected from the seven women?

$C_{7,4}=\frac{7!}{4!(7-4)!}=\frac{7!}{4!\times(3!)}=\frac{7\times6\times5\times\cancel{4!}}{\cancel{4!}\times3\times2\times1}=\frac{210}{6}=35$

c. Find the probability that the selected group will consist of all women.

4 women, from a total of 7: 35 combinations

(we calculated it in b.)

Total of Combinations: 4 people, from a total of 13: 715 combinations

(we calculated it in a.)

So, we have to write it as a fraction:

$\frac{35}{715}=\frac{7}{143}$