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Homework Help: Probability of a dice

  1. Jan 26, 2010 #1
    i throw a dice 5 times,
    1)what is the probability that i get the number 6 at least twice?

    2)what is the probability that i get the same number at least 4 times?


    i know that since i am throwing the dice 5 times i have 6^5=7776 options
    now all i need to find out is how many of those options contain 2 or more sixes, adn the same number 4 or more times
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    1) the number of options containing 6 twice

    i think it would be easier to find the amount that contain 0 or 1 six and subtract.
    since 0 or 1 of the rolls must be 6 and the rest must not be 6
    1*5*5*5*5 + 5*5*5*5*5 = 3750

    giving me the probability 3750/7776=0.482 that i will roll less than 2 sixes
    and the probability of 0.518 that i will roll 2 or more

    but the answer is meant to be 0.196?? where am i going wrong?
    had i used 5*5*5*5*5 + 5*5*5*5*5 =6250 instead of 3750 i would have gotten the right answer, but i doint understand why that is correct.

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    2) to get the same number 4 or more times i need to get the same number either 4 times or 5 times
    the first number can be any number, the next 3 must be the same as the first and the last must either be the same as the first or different to the first
    6*1*1*1*1 + 6*1*1*1*5=36

    36/7776=0.0046

    but again the correct answer is meant to be 0.02 and i cannot see how,
     
  2. jcsd
  3. Jan 26, 2010 #2

    vela

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    You didn't take into account the number of ways to roll a 6 only once. You have five possibilities: 6XXXX, X6XXX, XX6XX, XXX6X, XXXX6.

    You made the same error on the second problem.
     
  4. Jan 26, 2010 #3
    okay i think i got it

    5*5*5*5*5+(1*5*5*5*5)*5, and that really gives me 0.1962

    but for the second one

    (6*1*1*1*1)+(6*1*1*1*5) but do i multiply this by 5 also?? are there not more than 5 possibilities for the second term?
     
  5. Jan 26, 2010 #4
    also in another question which i thought i had right i was asked the possibility to roll the same number 3 times exactly out of 6


    i said

    6*1*1*5*5 since the first roll i can have any number, 2 others must be the same and 2 others must be different, but here too i didnt take into account the ways i could roll these numbers, i could have

    6 1 1 5 5
    1 1 5 5 6
    1 5 5 6 1
    5 5 6 1 1
    5 6 1 1 5
    1 6 1 5 5
    6 1 5 5 1
    6 1 ......................

    it seams like it never going to end, what os the method to solve this?
     
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