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. 1415
B. 45
C. 815
D. 715
E. 15
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