8 Volunteers for Charity Event

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bml1105 Really wants to Beat The GMAT!
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8 Volunteers for Charity Event Post Tue Jun 03, 2014 2:23 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post Tue Jun 03, 2014 2:35 pm
    Quote:
    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)
    = 2/7 = D

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

    Cheers,
    Brent

    Aside: If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789

    Aside: For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting?id=775

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    Post 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

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    Post 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.

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    osama_salah Just gettin' started! Default Avatar
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    Post 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 = 2/7 (D)

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