MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Find the sum of a probability of dice roll that is prime.Consider rolling n fair dice. Let p(n) be the probability that the product of the faces is prime when you roll n dice. For example, when n = 2, two dice are rolled and the probability that the product of the two dice is prime is p(2)= | { ( 1 , 2 ) ( 2 , 1 ) , ( 1 , 3 ) , ( 3 , 1 ) ( 1 , 5 ) ( 5 , 1 ) } | | 36 | = 1 6 When p ( 3 ) = 1 24 . Find the infinite sum of p(n) (when n starts at 0) (Hint: Consider differentiating both sides of the infinite geometric series: infinite sum of

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MathJax(?): Can&#039;t find handler for document MathJax(?): Can&#039;t find handler for document Find the sum of a probability of dice roll that is prime.Consider rolling n fair dice. Let p(n) be the probability that the product of the faces is prime when you roll n dice. For example, when   n  =  2, two dice are rolled and the probability that the product of the two dice is prime is p(2)=                    |            {      (      1      ,      2      )      (      2      ,      1      )      ,      (      1      ,      3      )      ,      (      3      ,      1      )      (      1      ,      5      )      (      5      ,      1      )      }              |                            |            36              |              =      1    6  When   p  (  3  )  =       1    24  . Find the infinite sum of p(n) (when n starts at 0) (Hint: Consider differentiating both sides of the infinite geometric series: infinite sum of

Addison Herman · Accepted Answer

Step 1You will get a prime product if exactly one of the faces is prime and the rest are all 1. The probability (binomial distribution) that you will have a prime product with n dice is                               (                            n        1                              )                      .    (          1      2              )      1        .    (          1      6              )              n        −        1            So you need to find             1      2              ∑              n        =        0            ∞        n    (          1      6              )              n        −        1            Replace       1    6   by x. Now consider the geometric sum   1  +  x  +      x    2    +      x    3    +  .  .  .      x    n    +  .  .  . hat happens when you differentiate it term by term wrt x? Especially consider what happens to the general term       x    n  .Step 2Now find an expression for the geometric sum in terms of x and differentiate that wrt x. Finally substitute   x  =      1    6  . That should give you the required sum.

Find the sum of a probability of dice roll that is prime.

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