mdecarbo wrote:From a group of 8 volunteers, including Adam and Karen, 4 are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 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 Karen
Take 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).
By 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 possible
We 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)
= [spoiler]2/7 = D[/spoiler]
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Cheers,
Brent
Aside: If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
