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Probability of two ones with 6 dice

  1. Feb 16, 2007 #1
    1. The problem statement, all variables and given/known data

    What is the probability of rolling exactly two ones from a set of six dice?

    2. Relevant equations
    Binomial probability distribution?


    3. The attempt at a solution
    I tried:

    P = 6!/(2! x 4!) x (1/6)^2 x (5/6)^4 and got 0.20, but the book says that's a factor of 10 too high!
     
  2. jcsd
  3. Feb 16, 2007 #2
    relevant equation is correct... but, let's look at it this way:

    You could get a 1,1,something else, something else, something else, something else.
    Probability is 1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6
    Or, you could get a
    something else, 1, something else, something else, 1, something else
    Probability is 5/6 * 1/6 * 5/6 * 5/6 * 1/6 * 5/6

    Note, the probabilities of these two are the same; after a bit of rearranging,
    (1/6)^2 * (5/6)^4

    Now, how many different orders can you get 2 ones and 4 something elses?
    The ones can be 1st and 2nd, 1st and 3rd, 1st and 4th, 1st and 5th, or 1st and 6th. Or, 2nd and 3rd, 2nd and 4th, etc. This is a combination of 2 elements out of 6, sometimes written as 6C2. (I don't know which way you've learned to write it.)

    So, put these two parts together, and you should get your answer.
     
  4. Feb 16, 2007 #3

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Whoever figured out the answers in your book is high on something. Your answer is correct.
     
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