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Probability

This topic has 4 expert replies and 0 member replies

Probability

Post Sat Sep 16, 2017 7:26 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

    A. 1/15
    B. 1/12
    C. 1/9
    D. 1/6
    E. 1/3

    OA is a

    why is E the only best option? which probability approach did you use to get your answer?[/spoiler]

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    Post Sun Sep 17, 2017 3:35 am
    Roland2rule wrote:
    Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

    A. 1/15
    B. 1/12
    C. 1/9
    D. 1/6
    E. 1/3
    Approach 1:
    From the 6 workers, the number of combinations of 2 that can be formed = 6C2 = (6*5)/(2*1) = 15.
    Of these 15 pairs, only 1 is good: Joshua and Jose.
    Thus:
    P(Joshua and Jose are both chosen) = 1/15.

    Approach 2:
    Joshua and Jose constitute 2 of the 6 workers.
    Thus:
    P(Joshua or Jose is chosen on the 1st pick) = 2/6.

    Since Joshua or Jose is chosen on the 1st pick, only 1 of the 5 remaining workers will be Joshua or Jose.
    Thus:
    P(Joshua or Jose is chosen on the 2nd pick) = 1/5.

    Since we want both events to happen, we multiply the probabilities:
    P(Joshua and Jose are both chosen) = 2/6 * 1/5 = 1/15.

    The correct answer is A.

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    Post Sun Sep 17, 2017 7:08 am
    Roland2rule wrote:
    Joshua and Jose work at an auto repair center with 4 other workers. For a survey on health care insurance, 2 of the 6 workers will be randomly chosen to be interviewed. What is the probability that Joshua and Jose will both be chosen?

    A. 1/15
    B. 1/12
    C. 1/9
    D. 1/6
    E. 1/3
    [/spoiler]
    P(Joshua and Jose both chosen) = P(one of the two men is chosen first AND the other man is chosen second)
    = P(one of the two men is chosen first) x P(the other man is chosen second)
    = 2/6 x 1/5
    = 1/15

    Answer: A

    Cheers,
    Brent

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    GMAT/MBA Expert

    Post Sun Sep 17, 2017 5:10 pm
    We could also do All Groups - Groups With Neither Guy - Groups With Only One of the Guys:

    All Groups = 6 choose 2 = 6!/(4!2!) = 15
    Neither Guy = 4 choose 2 = 4!/(2!2!) = 6
    Only One Guy = (4 choose 1) * (2 choose 1) = 8

    If Only One Guy doesn't make sense, I'm choosing one of the other four people first, then choosing one of our two guys to go with them.

    That leaves us with 15 - 6 - 8 = 1 group with both guys. 1 group out of 15 is 1/15, and we're done.

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    GMAT/MBA Expert

    Post Sun Sep 17, 2017 5:12 pm
    It seems like this is easy conceptually too. We want to pick both guys, so there's only one way to do that: 2 choose 2. If we're just picking two random guys, there are 6 choose 2, or 15 ways to do that, as shown in my previous post.

    From there:

    Our target = the one group
    Our total = 6 choose 2 = 15 groups

    Probability = target/total = 1/15

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