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Probability of receiving bonus

  1. Jul 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A factory of 36 male workers and 64 female workers. woth 10 male workers earning less than $1000 per month and 17 female workers earning at least $1000 per month . At the end of year, workers earning less than $1000 are given a bonus of $1000 whereas others receive a month's salary.
    If a male worker and a female worker are randomly chosen , find the probability that exactly one worker receive one month salary
    here's my working.

    P(male receive one month salary , female not receive one month salary) +P(female not receive one month salary , male receive one month salary) + P(female receive one month salary , male not receive one month salary) +P( male not receive one month salary , female receive one month salary

    = (0.26 x (47/99)x2 ) + ( (0.17x (10/99) x2 ) = 0.28

    but the ans is 0.604
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2014 #2

    HallsofIvy

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    You are making the same mistake you did in the almost identical problem you posted with "RM" rather than "$". You are using "100" as the denominator when you are given that one person is a male and the other female. Use "36" as the denominator for the male and 64 for the female worker.

    The probability that a male worker earns less than $1000 is 10/36 and over $1000 is 26/36.

    The probability that a female worker earns less than $1000 is 17/64 and over $1000 is 47/64.
     
  4. Jul 25, 2014 #3

    Ray Vickson

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    There are two possible interpretations:
    (1) One male chosen at random from the male workers and one female chosen at random from the female workers.
    (2) Two workers chosen randomly; one of them happens to be male and one female.

    I suspect that (1) is the intended interpretation, but your working seems to be for (2). How would you tackle (1)?

    Note: the two interpretations could give very different answers.
     
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