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Math Help - need help to confirm my probability understanding

  1. #1
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    need help to confirm my probability understanding

    I have 4 unbiased dice with 6 sides.

    1) I want dice1 to have value "1" and any values from the other 3 dice.
    Ans: 1/6 x 1 x 1 x 1 = 1/6

    2) I want at least 3 dice to have values "1"
    Ans: 4c3 x 1/6 x 1/6 x 1 x 1 = 1/9

    3) I want at least 3 dice with the same values (any value as long as it's the same)
    Ans: 4c3 x 1/6 = 2/3


    Need some help to check if my understanding of probability is correct. It has been a long time since I worked on this topic. I think I might be wrong with no.3 or if I'm wrong with any, please advise.

    Thanks!
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  2. #2
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    Quote Originally Posted by zepher View Post
    I have 4 unbiased dice with 6 sides.

    1) I want dice1 to have value "1" and any values from the other 3 dice.
    Ans: 1/6 x 1 x 1 x 1 = 1/6

    2) I want at least 3 dice to have values "1"
    Ans: 4c3 x 1/6 x 1/6 x 1 x 1 = 1/9
    No, for exactly three of them show a "1" so the probability is:

    4c3 (1/6)^3 = 1/56

    and for four of then to show 1 the probability is:

    4c4 (1/6)^4=(1/6)^4

    So the probability of three or more 1's is:

    4c3 (1/6)^3 + (1/6)^4

    3) I want at least 3 dice with the same values (any value as long as it's the same)
    Ans: 4c3 x 1/6 = 2/3
    No, we gave the probability that three or more show a specified face in 2 above, but
    there are 6 possible specifications, so this is 6 times the answer to 2.

    RonL
    Last edited by CaptainBlack; April 19th 2008 at 10:51 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    No, three of them show a "1" so the probability is:

    4c3 (1/6)^3 = 1/56

    RonL
    Uhhh...

    Shouldn't that be 4c3 (1/6)^3 + 4c4 (1/6)^4?

    The original post asked for the probability of at least three 1's-- in other words, three 1's or four 1's.
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  4. #4
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    Quote Originally Posted by awkward View Post
    Uhhh...

    Shouldn't that be 4c3 (1/6)^3 + 4c4 (1/6)^4?

    The original post asked for the probability of at least three 1's-- in other words, three 1's or four 1's.
    Opps, yes. Post corrected.

    RonL
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  5. #5
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    thanks a bunch!
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