Asked by Raj
Four fair sided dice, each with faces marked 1,2,3,4,5,6, are thrown on 7 occasions.Find the probability that the numbers shown on the four dice add up to 5 on exactly one or two of the 7 occasions.
Answers
Answered by
R_Scott
there are 6^4 possible outcomes with 4 dice
there are 4 ways to get a sum of 5
... 3 dice show one, and the 4th shows 2
f = p(5) = 4 / (6^4) ... n = p(not 5) = 1 - [4 / (6^4)]
this is binomial , 5 or (not 5) over 7 throws
(n + f)^7 = n^7 + 7 n^6 f + 21 n^5 f^2 + ... + 7 n f^6 + f^7
the answer is the sum of the 2nd and 3rd terms
there are 4 ways to get a sum of 5
... 3 dice show one, and the 4th shows 2
f = p(5) = 4 / (6^4) ... n = p(not 5) = 1 - [4 / (6^4)]
this is binomial , 5 or (not 5) over 7 throws
(n + f)^7 = n^7 + 7 n^6 f + 21 n^5 f^2 + ... + 7 n f^6 + f^7
the answer is the sum of the 2nd and 3rd terms
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