Question 430186
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Hi
P(11 with one roll of 2 dice) = 2/36 = 1/18 = .0556  P(not 11) = .9444
___1_2_3__4__5__6
1|_2_3_4__5__6__7
2|_3_4_5__6__7__8
3|_4_5_6__7__8__9
4|_5_6_7__8__9_10
5|_6_7_8__9_10_11
6|_7_8_9_10_11_12
P(7 rolls, sum of 11 exactly 3 times) = 7C3(.0556)^3(.9444)^4 
Note: The probability of x successes in n trials is: 
P = nCx* {{{p^x*q^(n-x)}}} where p and q are the probabilities of success and failure respectively. 
In this case p = .0556 & q =.9444
nCx = {{{n!/(x!(n-x)!)}}}