Carcass wrote:
A club has exactly 3 men and 7 women as members. If two members are selected at random to be president and vice-president respectively, and if no member can hold two offices simultaneously, what is the probability that a woman is selected for at least one of the positions?
A. \(\frac{14}{15}\)
B. \(\frac{4}{5}\)
C. \(\frac{8}{15}\)
D. \(\frac{7}{15}\)
E. \(\frac{1}{5}\)
We can solve this by using counting methods or by applying probability rules.
Let's use probability rules
We want P(selecting
at least 1 woman)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A
not happening)
So, here we get: P(selecting at least 1 woman) = 1 -
P(not selecting at least 1 woman)What does it mean to
not select at least 1 woman? It means selecting ZERO women.
So, we can write: P(selecting at least 1 woman) = 1 -
P(selecting ZERO women)P(selecting ZERO women) = P(selecting TWO men)
= P(selecting a man to be president
AND selecting a man to be vice-president)
= P(selecting a man to be president)
x P(selecting a man to be vice-president)
= 3/10
x 2/9
= 6/90
=
1/15ASIDE: For the 1st selection, 3 of the 10 people are men.
Once we select a man for the 1st selection, there are 9 people remaining, and 2 of them are men.
So, P(selecting at least 1 woman) = 1 -
1/15= 14/15
Answer:
ACheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep