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Counting question

  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A box of valentine candy hearts contains 52 hearts of which 19 are white, 10 are tan, 7 are pink, 3 are purple, 5 are yellow, 2 are orange and 6 are green. If you select 9 pieces at random (without replacing any), what is the probability that 3 are white, 2 are tan, 1 is pink, 1 is yellow and 2 are green?

    2. Relevant equations
    (n/r) = n! / r!(n-r)!

    3. The attempt at a solution
    So you select 9 pieces at random... you want to draw 3 that are white and there at 19 white total out of 52 possible, so the chance of getting a white on your first draw is 19/52, then 18/51, then 17/50,

    for 2 tan out of 10 tan with 50 remaining, i'd go 9/49 then 8/48.

    I would continue this process for all colors then multiply these fractions together, is this correct? if not can somebody give me some insight?
     
  2. jcsd
  3. Feb 25, 2016 #2

    Ray Vickson

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    Yes, you would multiply them together; but you would not be finished. The three whites need not be the first three candies drawn; perhaps the first drawn white is candy #4, the second might be candy #7 and the last candy #9. Similar ordering issues occur for all the other colors as well, so you have a non-trivial combinatorial counting problem to enumerate all the different strings of size 9, consisting of 3 Ws, 2 Ts, etc.

    Start with a simpler case: 2 colors (C1 and C2), with N1 items of type C1 and N2 of type C2 (N = N1+N2 total). The probabilty of drawing k1 items of type C1 (and, of course, n-k1 of type C2) in n draws (without replacement) has a well-known form called the hypergeometric distribution, and can be found in any good probability textbook, or on-line.

    Now, of course, you have 7 types instead of 2, but you can do it sequentially: The probability of getting 3 white and 6 non-white is given by a simple hypergeometric with parameters N1=19, N2 = 52-19 = 33, n = 9 and you want k1 = 3 . Now, among the 6 non-white the probability of getting 2 tan and 4 not (tan or white) is hypergeometric with parameters N1 = 10, N2= 33-10 = 23, n = 6 and you want k1 = 2, etc. You can safely multiply all those probabilities together without worrying about the order in which the colors are drawn, because that aspect has already been taken care of when you employ the hypergeometric distribution.
     
  4. Feb 25, 2016 #3

    haruspex

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    It won't hurt to read up on distributions in textbooks, but it is not really necessary here.
    How many ways are there altogether of choosing 9 from the 52?
    How many ways are there of choosing 3 of the 19 white, and two of the ten tan, and .... etc.
     
  5. Mar 8, 2016 #4
    Well, the problem doesn't state that the desired elements have to be drawn in a sequence, so you don't have to actually get the first three white hearts in the first three draws. Think about your formula.
     
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