8 Volunteers for Charity Event

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bml1105: Master | Next Rank: 500 Posts; Posts: 113; Joined: Sun Nov 24, 2013 9:19 pmFollowed by: 5 members; Thanked: 1 times

by bml1105 » Tue Jun 03, 2014 2:23 pm

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

OA: D

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Source: Beat The GMAT — Problem Solving | Tue Jun 03, 2014 2:23 pm

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by Brent@GMATPrepNow » Tue Jun 03, 2014 2:35 pm

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]

------------------------------

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

Brent Hanneson - Creator of GMATPrepNow.com

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by [email protected] » Wed Jun 04, 2014 12:06 am

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

Contact Rich at [email protected]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

Adam not Karen

by GMATGuruNY » Wed Jun 04, 2014 1:50 am

bml1105 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

ONE WAY:
One way to yield a favorable outcome is to choose Aaron first.
P(1st person selected is Aaron) = 1/8.
P(2nd person selected is not Karen) = 6/7. (Of the 7 remaining people, anyone but Karen.)
P(3rd person selected is not Karen) = 5/6. (Of the 6 remaining people, anyone but Karen.)
P(4th person selected is not Karen) = 4/5. (Of the 5 remaining people, anyone but Karen.)
To combine these probabilities, we multiply:
1/8 * 6/7 * 5/6 * 4/5 = 1/14.

OTHER WAYS:
Since a favorable outcome will be achieved if Aaron is chosen 1st, 2nd, 3rd, or 4th -- for a total of 4 ways -- we multiply by 4:
1/14 * 4 = 2/7.

The correct answer is D.

Last edited by GMATGuruNY on Mon Jan 08, 2018 3:40 am, edited 1 time in total.

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osama_salah: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Wed Jun 15, 2016 2:05 am

by osama_salah » Fri Jul 08, 2016 10:15 pm

An old post but I have a straightforward solution:

P(Andrew is selected AND Karen not selected) =
[1 - P(Andrew not selected in one of the 4 selections)] * P(Karen not selected in the subsequent 3 selections)

= [1 - (7/8 * 6/7 * 5/6 * 4/5)] * (6/7 * 5/6 * 4/5)
= [1 - 4/8] * 4/7
= 1/2 * 4/7 = [spoiler]2/7 (D)[/spoiler]

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