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  • another urn question

    A bag contains 3 red balls, 2 white balls and 3 blue balls. Three balls are selected randomly from the bag with replacement. Given that no blue ball has been selected, calculate the probability that the number of red balls selected exceeds the number of white balls.

    ans choices: (a) 3/8 (b) 3/5 (c) 7/10 (d) 81/512 (e) 81/125


    My solution: Want P(R>W|B'), where R = # of red, W = # of white, B = # of blue = P(R>W and B')/ P(B')
    = [(3/8)^2(2/8)*3!/2!1! + (3/8)^3] / (1-(3/8)^3)
    = not the answer given above.

    reasoning: (3/8)^2 * 2/8 multiplied by 3!/2! since this is the probability of getting two red and 1 white, added with 3/8^3 since this is the probability of choosing three red.

    The denominator is due to the fact that to pick no blue is to find 1-prob of choosing all three blue

    Thank you

  • #2
    I'll try to take a look at it when I get home.
    act justly. walk humbly. .

    Comment


    • #3
      Are you sure about the answer choices? I keep getting 81/117 for this problem.

      Anyway, the problem with your solution is the denominator-

      P (no blue) is not equal to 1 - P(all blue). Think about this!

      P (no blue) = 1 - P(atleast 1 blue) = 1 - [P (1blue/2blue/3blue)].

      Please post the answer to this problem.

      Comment


      • #4
        Ok, I got (e) again on this one. First, I think it would be helpful to be somewhat familiar with multinomials involving the choosing of more than one element at a time, to which magic was referring in the last problem. Essentially, if we have X balls of two types, and we would like to choose m of type 1, and n of type 2, where m+n = X, then

        "X choose m and n" = X!/(m!n!)

        Now, let's see what combinations of 3 drawn balls of ONLY reds and whites will give us more reds than whites (this is done since NO blue balls are drawn):

        3 R, 0 W; 2 R, 1 W.

        Now, the possible combinations of 3 drawn balls of ONLY reds and whites are:

        3 R, 0 W; 2R , 1 W; 1 R, 2 W; 0 R, 3 W.

        Now, the number of ways we can draw 3 R, 0 W with replacement is:

        (3 * 3 * 3) * ("3 choose 3 and 0") = 27 * 3!/(3!*0!) = 27

        If you understand this, then the rest follows quite directly:

        # for 2R, 1W: (3 * 3 * 2) * ("3 choose 2 and 1") = 18 * 3!/2!1! = 54
        # for 1R, 2W: (3 * 2 * 2) * ("3 choose 1 and 2") = 36
        # for 0R, 3W: (2 * 2 * 2) * ("3 choose 0 and 3") = 8

        So, Prob( R > W | B = 0) = (27+ 54) / (27 + 54 + 36 + 8) = 81 / 125

        This is another way to look at the problem instead of framing everything in terms of probabilities. Maybe it will be less confusing, maybe not. Regardless, though, if this answer agrees with the book, then hopefully seeing the problem attacked like this will help.
        Last edited by .Godspeed.; June 20 2005, 07:21 PM.
        act justly. walk humbly. .

        Comment

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