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Counting/Probability Problems

  1. Apr 4, 2009 #1
    Hello:

    I am wondering if I am doing the following correctly. This is the problem:

    (a) how many different outcomes are possible when a pair of dice are rolled 2 successive times?

    roll once with 2 dice, 36, so roll twice, 36*36.

    (b) what's the probability that each die shows same value on second roll as first row?

    for 1 dice, there are 6 cases of the same value, so it's 1/6. For both die, it's (1/6)*(1/6) = 1/36.

    (c) what's the probability that the sume of the two dice is the same on both rolls?

    For this you have the following possibilities:

    (1 1) (2 2) (3 3) (4 4) (5 5) (6 6)
    (1 2) (1 3) (1 4) (1 5) (1 6)
    (2 3 ) (2 4) (2 5) (2 6)
    (3 4) (3 5) (3 6)
    (4 5) (4 6)
    (5 6)

    The possible sums are : 2 (1 1)
    4: (2 2) (1 3)
    5: (1 4) (2 3)
    6: (3 3) (1 5) (2 4)
    7: (3 4) (2 5)
    8: (4 4) (2 6) (3 5)
    9: (3 6) (4 5)
    10: (5 5) (4 6)
    11: (5 6)
    12 (6 6)

    So, the probability of rolling a (1 1) both times is (1/6)^4.
    Rolling a 4 both times is (1/6)^4 + ((1/6)^4))*2 (because you can have 1 3 or 3 1)
    Rolling a 5 is (1/6)^4 * (2) * 2
    and so on?
     
  2. jcsd
  3. Apr 5, 2009 #2

    tiny-tim

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    Hello bodensee9! :smile:

    (a) and (b) look good.
    That's basically the correct method …

    but what happened to 2 2?
     
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