Carcass wrote:
From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?
A. 3/7
B. 5/12
C. 27/70
D. 2/7
E. 9/35
P(Andrew is selected but Karen is not selected) = (
number of 4-person groups with Andrew but not Karen)/(
total # of 4-person groups possible)
number of 4-person groups with Andrew but not KarenTake the task of creating groups and
break it into stages.
Stage 1: Place Andrew in the 4-person group
We can complete this stage in
1 way
Stage 2: Send Karen out of the room and, from the remaining 6 volunteers, select 3 more people.
Since the order in which we select the 3 volunteers does not matter, we can use combinations.
We can select 3 people from 6 volunteers in 6C3 ways (
20 ways).
the video below shows you how to calculate combinations (like 6C3) in your headBy the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a 4-person group) in
(1)(20) ways (=
20 ways)
total # of 4-person groups possibleWe can select 4 people from all 8 volunteers in 8C4 ways ( =
70 ways).
So, P(Andrew is selected but Karen is not selected) = (
20)/(
70)
= 2/7
Answer: D
Cheers,
Brent