SOLUTION: Can I get help with this problem please: In 7 rolls of 2 fair dice what is the probabilty of obtaining a sum of 11 exactly 3 times? Thanks, Kim

 

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*  where p and q are the probabilities of success and failure respectively. 

In this case p = .0556 & q =.9444

nCx =