BTGmoderatorLU wrote: ↑

Fri Aug 28, 2020 10:03 am

Source: Princeton Review

There are 8 job applicants sitting in a waiting room, 4 women and 4 men. If 2 of the applicants are selected at random, what is the probability that both will be women?

A. \(\dfrac{1}{4}\)

B. \(\dfrac{3}{7}\)

C. \(\dfrac{5}{2}\)

D. \(\dfrac{3}{14}\)

E. \(\dfrac{1}{10}\)

The OA is D

P(both selected people are women) = P(1st selection is a woman AND 2nd selection is a woman)
= P(1st selection is a woman) x P(2nd selection is a woman)
= 4/8 x 3/7
= 3/14

Answer: D

Aside:
P(1st selection is a woman) = 4/8, since there are 8 people, and 4 of them are women
P(2nd selection is a woman) = 3/7. Once we have selected a woman for the 1st selection, there are 7 people remaining, and 3 of them are women

Cheers,
Brent

BTGmoderatorLU wrote: ↑

Fri Aug 28, 2020 10:03 am

Source: Princeton Review

There are 8 job applicants sitting in a waiting room, 4 women and 4 men. If 2 of the applicants are selected at random, what is the probability that both will be women?

A. \(\dfrac{1}{4}\)

B. \(\dfrac{3}{7}\)

C. \(\dfrac{5}{2}\)

D. \(\dfrac{3}{14}\)

E. \(\dfrac{1}{10}\)

The OA is D

We can also solve this question using counting methods.

P(both selections are women) = (# of outcomes where 2 WOMEN are selected)/(TOTAL # of possible outcomes)

TOTAL # of possible outcomes
Since the order in which we select the two people does not matter, we can use combinations.
We can select 2 people from 8 people in 8C2 ways
8C2 = 28


# of outcomes where 2 WOMEN are selected
Since the order in which we select the two women does not matter, we can use combinations.
We can select 2 women from 4 women in 4C2 ways
4C2 = 6


So, P(both selections are women) = 6/28
= 3/14

Answer: D