Problem Solving

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

Hi bml1105,

There's another way to "do the math" behind this question, although it takes a little longer - we can treat this question as a straight probability question and keep track of ALL the possible ways to get "what we want"....

We have 8 people to choose from; we'll pick 4 - we WANT Andrew in our group, we DON'T WANT Karen in our group.

_ _ _ _

If we want Andrew to be the first person we pick, then the other 3 CAN'T be Karen. We end up with the following math (notice that after picking each person, the total number of people remaining decreases):

(1/8)(6/7)(5/6)(4/5) = 4/56 = 1/14

Andrew COULD be the second, third or fourth person chosen though. Here's the math if he's SECOND and we don't pick Karen...

(6/8)(1/7)(5/6)(4/5) = 4/56 = 1/14

Notice how the product is the SAME.

If he's third, it would still be 1/14
if he's fourth, it would still be 1/14

With ALL of those possibilities, we end up with 1/14 + 1/14 +1/14 + 1/14 = 4/14 = 2/7

Final Answer: D

GMAT assassins aren't born, they're made,
Rich