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Math Help - probability of receiving bonus

  1. #1
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    probability of receiving bonus

    A factory of 36 male workers and 64 female workers. worth 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
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  2. #2
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    Re: probability of receiving bonus

    here's another part of this question,

    find the probability of the three doctors selected . the correct working would be (20C3 X15C1)/35C4 = 0.327

    can i do in this way? P(DDDE) + P(EDDD) + P(DEDD) +P(DDED) =
    ( (20/35) x (19/34) x (18/35) x (15/32) ) x 4 = 0.327

    Is my concept wrong? D=doctor E=engineer
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  3. #3
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    Re: probability of receiving bonus

    Quote Originally Posted by delso View Post
    A factory of 36 male workers and 64 female workers. worth 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
    fixed \$ signs
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  4. #4
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    Re: probability of receiving bonus

    Quote Originally Posted by romsek View Post
    fixed \$ signs
    find the probability of the three doctors selected . the correct working would be (20C3 X15C1)/35C4 = 0.327

    can i do in this way? P(DDDE) + P(EDDD) + P(DEDD) +P(DDED) =
    ( (20/35) x (19/34) x (18/35) x (15/32) ) x 4 = 0.327

    Is my concept wrong? D=doctor E=engineer
    Follow Math Help Forum on Facebook and Google+

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