When three standard dice are tossed, the numbers a,b,c are obtained. Find the probability that (a-1)(b-1)(c-1) not equal to 0
+93 the only possible way (a-1)(b-1)(c-1) is not zero is if none of the dice have "1" on the face. There are the ways a=1 b=1 c=1 a=b=1 a=c=1 b=c=1 or a=b=c=1, and there are 216 total possible outcomes, so the answer is 216-7/216=209/207. (I may not have simplified)
