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Simple probability question

  1. Sep 2, 2011 #1
    I got lost with simple probability question.
    There are three varaeties of candies in a can.
    Three candies are selected randomly.
    What is the probability that all of them are of the same kind?

    It seems to be [itex]1\frac{1}{3}\frac{1}{3}=\frac{1}{9}[/itex]

    But at the same time we can calculate it as [itex]\frac{C^1_3}{C^3_{3+3-1}}=\frac{3}{10}[/itex]

    What is the right answer here?
  2. jcsd
  3. Sep 2, 2011 #2
    You didn't specify the proportion of the three kinds of candies in the can. You can estimate that by sampling with replacement, but that's a different question. For this question let's say each of the three kinds are equal in number. Let [itex]n_i[/itex] be the number of each kind and the total [itex] N=\sum_{i=1}^{i=3}[/itex]

    The first draw determines the target kind and is drawn with probability one. What is the probability of the same kind at the next draw without replacement? If you've drawn the same kind on the second draw, what is the probability of drawing the same kind on the third draw without replacement?

    It's clear that if you draw with replacement, the probability is 1/9. so your calculation cannot be correct for drawing without replacement.
  4. Sep 3, 2011 #3
    Thank you for response.
    Let say candies are at the same proportion and this is draw with replacement (we return candy to the can after every draw).

    The issue is that if I try to calculate number of possible outcomes I get


    We can enumerate them as

    v1 | v2 | v3
    0 0 3
    0 1 2
    0 2 1
    0 3 0
    1 2 0
    2 1 0
    3 0 0
    2 0 1
    1 0 2
    1 1 1

    Each outcome is equally likely so based on this enumeration probability is [itex]\frac{3}{10}[/itex]

    I expect it to be [itex]\frac{1}{9}[/itex]

    This is stupid but I do not see where I am wrong here :-(.
  5. Sep 3, 2011 #4
    Well, first the 'with replacement' problem is not very interesting or practical. Who takes candy from a jar just to put it back? However if you happen to get your favorite flavor with the first draw, it might be interesting know what the probabilities of getting more like that.

    You are dealing with only two columns, not three. The first draw simply establishes the "target". The draws are independent, so it's simply 1/3 probability for each of the remaining draws. To get the target flavor twice in a row would simply be the product or 1/9. If you started with a preconceived choice of flavor, then the probability of getting three in a row of that flavor is of course 1/27 with replacement.

    However without replacement you need to adjust the denominator for the items removed. If you are only interested in getting three in a row after establishing your target with the first draw, you must calculate the new proportion for the second and third draw given your series is intact.:

    [itex] P_2 = n_i - 1/N - 1;, P_3 = n_i -2/N - 2[/itex] I don't see this as a combinatorial problem. The product of P2 and P3 is your answer since P1=1.

    EDIT: You would use a combinatorial approach if you asked about how many ways you can draw n things from N things in a non contingent context If you want to ask that question, you need to re post.
    Last edited: Sep 3, 2011
  6. Sep 3, 2011 #5
    Each outcome is NOT equally likely when listed as above.

    P(3,0,0) = (1/3)^3

    P(1,1,1) = 6 x (1/3)^3
  7. Sep 3, 2011 #6
    Well, I can re-frase same problem in 100 different ways. Let say we are talking about three guys randomly selecting from three targets to shoot to. Does it make it more interesting? :-). I would say exactly the oposite. Same problem without replacement it totally trivial and not interesting.

    I totally agree, that is why in my first post I said that it seems to be [itex] 1\frac{1}{3}\frac{1}{3} [/itex].
    However, in this case we have finite number of equally likely outcomes (unless you can see why they are not equally likely).
    As such we can calculate probability as [itex]\frac{\text{number of favorable outcomes}}{\text{number of possible outcomes}} [/itex]

    So we should be able to tackle same problem from combinatorial prospective.

    The issue is that the answer I am getting using this approach is [itex]\frac{3}{10}[/itex].

    So I am trying to figure out what I am doing wrong.
    Last edited: Sep 3, 2011
  8. Sep 3, 2011 #7
    Thanks awkward.

    Finally I got it. I thought I am totally crazy :-).

    [itex] p(1,1,1) = 1(1-\frac{1}{3})(1-\frac{2}{3}) =\frac{2}{9} [/itex]
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