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Probability Question

  1. Feb 17, 2010 #1
    1. Question http://tinyurl.com/ydwpqx4

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    Hello. The link I provided has the entire statement of the question, and it's solution (Example 4.8).
    We know that P(1 or 6) for one throw = 1/3, and P(neither 1 nor 6) = 2/3.
    What I don't understand is why we don't cube 1/3 to get the probability of getting 1 or 6 for 3 die. The solution first cubes 2/3 and then subtracts it from the total probability 1. If we can get the probability of not getting 1 or 6 by cubing it's probability, why can't we get the the probability of getting 1 or 6 by cubing it's probability?

    Thanks.
     
  2. jcsd
  3. Feb 17, 2010 #2

    CompuChip

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    (1/3)3 is the probability that you get 1 or 6, on all three throws.

    What you want is the probability that you get 1 or 6 on at least one of them. So you could write out the list (where Y means throwing 1 or 6, and N means any other number):
    P(1 or 6 shows up) = P(first throw gives one or 6) + P(second throw gives 1 or 6) + P(third throw gives 1 or 6) = P(Y) + P(NY) + P(NNY) = (1/3) + (2/3)(1/3) + (2/3)(2/3)(1/3) = (1/3)(1 + 2/3 + 4/9) = (1/3)(19/9) = 19/27.

    The solutions take a shortcut, using the complement rule.
     
  4. Feb 17, 2010 #3
    Thanks a lot for your reply, and you cleared up the problem for me. But...

    This is where I lose you. Wouldn't the probability of getting 1 or 6 in the second throw be 1/3 as well?
     
  5. Feb 17, 2010 #4

    CompuChip

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    Yes.

    But what I mean by P(second throw gives 1 or 6) is actually: P(second throw gives 1 or 6 and first throw doesn't).
    Since the throws are independent of one another, this is

    P(second throw gives 1 or 6) x P(first throw doesn't give 1 or 6) = (1/3) x (2/3).

    This counts the possibilities YY, YN (from the first 1/3) and NY (from the 1/3 x 2/3).

    Just writing 1/3 also includes the possibility of both the first and the second giving 1 or 6, and you would count: YY, YN (from the first 1/3) and NY, YY (from the second 1/3). You see that you are double-counting YY (1 or 6 in both throws).
     
  6. Feb 17, 2010 #5
    I get it now. Thanks a lot, you're a life saver!
     
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