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Posted by on Tuesday, August 5, 2008 at 8:38pm.

A golf bag contains 6 white balls and 8 yellow balls. What is the probability of each event?

A) Drawing 3 white balls?

B.) Drawing 1 Yellow ball and 1 white ball?

Thanks

  • MATH 12 - , Tuesday, August 5, 2008 at 9:33pm

    in A, are you drawing only 3 balls?
    then prob = (6/14)(5/13)4/12) = 5/91

    or C(6,3)/C(14,3) = 20/364 = 5/91

    in B, are you drawing only 2 balls?
    then (8/14)(7/13) = 4/13
    or C(8,2)/C(14,2) = 28/91 = 4/13

  • MATH 12 - , Tuesday, August 5, 2008 at 9:42pm

    binomial distribution
    p white = 6/14 = 3/7
    p yellow = 1 - pwhite = 4/7

    in three trials, what is the chance of drawing exactly 3 white?
    The binomial coef of 3 out of three is 1
    because 3!/(3!(3-3)!) = 1
    P(3/3) = 1 * (3/7)^3 * (4/7)^0
    so (3/7)^3

    What is the probability of drawing exactly one white out of 2 trials?
    C(2/1) = 2!/[1! (2-1)!] = 2
    P(1/2) = 2 * (3/7)^1 * (4/7)^1
    = 24/49

  • MATH 12 - , Wednesday, August 6, 2008 at 9:25am

    Use what Reiny wrote. I ignored the changes in probability as the bag contained fewer balls.

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