Carcass wrote:
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. \(\frac{1}{4}\)
B. \(\frac{3}{7}\)
C. \(\frac{5}{2}\)
D. \(\frac{3}{14}\)
E. \(\frac{1}{10}\)
APPROACH #1: Counting methodsP(both selections are women) = (
# of outcomes where 2 WOMEN are selected)/(
TOTAL # of possible outcomes)
TOTAL # of possible outcomesSince 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 selectedSince 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 =
6So, P(both selections are women) =
6/
28= 3/14
Answer: D
APPROACH #2: Probability rulesP(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