Gmat_mission wrote: ↑

Thu Sep 10, 2020 12:14 am

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. \(\dfrac14\)

B. \(\dfrac37\)

C. \(\dfrac52\)

D. \(\dfrac3{14}\)

E. \(\dfrac1{10}\)

Answer: D

Source: Princeton Review

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

Gmat_mission wrote: ↑

Thu Sep 10, 2020 12:14 am

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. \(\dfrac14\)

B. \(\dfrac37\)

C. \(\dfrac52\)

D. \(\dfrac3{14}\)

E. \(\dfrac1{10}\)

Answer: D

Solution:

The number of ways to choose 2 applicants from 8 is 8C2 = (8 x 7)/2 = 28.

The number of ways to choose 2 female applicants from 4 is 4C2 = (4 x 3)/2 = 6.

Therefore, the probability that both applicants are women is 6/28 = 3/14.

Alternate Solution:

The probability that the first selection is a woman is 4/8 = 1/2. Assuming a woman is already selected, the probability that the second selection is a woman is 3/7. Thus, the probability that both selections are women is 1/2 * 3/7 = 3/14.

Answer: D