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Tricky urn problem

  1. Apr 5, 2008 #1
    Would anyone kindly be able to show me how to solve a problem like this?

    Suppose there are 8 white balls, 7 black balls and 4 red balls in an urn. If three balls are selected from the urn without replacement, what is the probability that two of them only are red?
  2. jcsd
  3. Apr 5, 2008 #2
  4. Apr 6, 2008 #3
    Ah thank you very much!

    Could you please help me understand how you figured it out though?
    I'm trying to follow your working but it doesn't seem to add up for me..
  5. Apr 6, 2008 #4
    Plymouth calculated the # of ways to choose 2 red out of 4 possible red, times the # of ways to choose 1 not red out of 15 possible not red. Then divided by the total possible outcomes.
  6. Apr 6, 2008 #5
    Oh ok.. so is it just me though or does the first part of the equation not equal the second part.. ie 0.21 vs 0.093?
  7. Apr 6, 2008 #6

    D H

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    You violated one of the sites rules. Please do not give out answers. Our goal here is to help people learn, and giving out answers does not accomplish that goal.


    The first thing to do is to get rid of superfluous information. All you care about here is red balls versus non-red balls. That some non-red balls are white and some are black is superfluous to this problem. You have 4 red balls and 15 non-red balls in the urn.

    Sometimes it helps to look at these kinds of problems from a slightly different point of view. You are asked to find the probability that exactly two out of three of the balls drawn from the urn is red. That is exactly the same as the probability that exactly one of the three balls drawn from the urn is not red. The problem is much easier to solve (for me, at least) when looked at this way. You can draw the non-red ball on the first draw, the second, or the third. These are three mutually exclusive events (non-red, red, red versus red, non-red, red versus red, red, non-red).
    • Do these three distinct events have equal probabilities?
    • What is the probability of drawing a non-red ball, a red ball, and another red ball in that order?
    • Given the answers to the above questions, what is the probability of drawing a non-red ball, a red ball, and another red ball in any order?
  8. Apr 7, 2008 #7
    Ah I have it now!

    The probability of three separate cases = the sum of the probability of each case

    --> R R NR = 4/19 x 3/18 x 15/17 = 10/323

    --> R NR R = 4/19 x 15/18 x 3/17 = 10/323

    --> NR R R = 15/19 x 4/18 x 3/17 = 10/323

    10/323 + 10/323 + 10/323 = 30/323 = 0.093

    Thanks guys, and thanks also to Gib Z who gave me some good advice too.
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