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Probability problem -- Number of throws of a pair of dice to get a 7

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data
    The minimum number of throws of a pair of dice so that the probability of getting the sum of the digits on the dice equal to 7 on atleast one throw is greater than 0.95, is n. Find n.

    2. Relevant equations


    3. The attempt at a solution
    There are 6 possible ways of getting a sum of 7.
    So, P(7) = 6/36 = 1/6 and P(not 7) = 5/6
    For minimum number of throws, on the last throw the sum should be 7.
    So,
    (5/6)n-1 (1/6) > 0.95
    (5/6)n-1 > 5.7

    Solving for n gives an incorrect answer.
     
  2. jcsd
  3. Aug 11, 2015 #2

    Orodruin

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    You are computing the probability of rolling 7 on the nth roll. Not the probability of obtaining at least one seven in n rolls.
     
  4. Aug 11, 2015 #3
    But wouldn't the minimum number of throws correspond to a 7 on the last throw?
     
  5. Aug 11, 2015 #4
    Doing what you said does give the right answer--
    1- (5/6)n < 0.95

    n= 17

    But I don't understand
     
  6. Aug 11, 2015 #5

    Orodruin

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    They are not asking for having a 7 on the last throw. They are asking how many throws you have to make to have a 95% chance of getting at least one seven, the seven can be in any of those throws (or several).
     
  7. Aug 11, 2015 #6
    Okey got it! Thanks :)
     
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