mdecarbo wrote:From a group of 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

I think the question is missing some key information.
How many volunteers are there altogether?

Cheers,
Brent

Hey Mike,

I hope you're doing well.
You might want to go back and edit your first post, so that others will see the complete question.

Cheers,
Brent

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

feedrom wrote:

Brent@GMATPrepNow wrote: 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).

Hi Brent,
In case if we have to choose 4 people include Karen, the desired out come will be 6C2 = 15. Is that right?

Exactly.
We automatically place Karen and Adam on the "team," and then select 2 more people from the remaining 6 people.

Cheers,
Brent

parveen110 wrote:Hi Brent!!

Could you please also give an alternate solution where order matters. I'm getting confused when trying to solve the other way.

Or what is wrong with my approach:

# of ways we may select remaining three persons= 6 x 5 x 4

# of ways we may select all four out of eight = 8 x 7 x 6 x 5

Probability of required selection = (6 x 5 x 4)/(8 x 7 x 6 x 5)= 1/14

Thank you!

You almost have it.
If we're saying that order matters, we need to remember that Adam can be selected 1st, 2nd, 3rd, or 4th.
So, for each of the (6 x 5 x 4) selections that include Adam, we can select Adam in 4 ways.

So, Probability of required selection = (6 x 5 x 4 x 4)/(8 x 7 x 6 x 5) = 2/7

Cheers,
Brent