Suppose a pair of fair dice is tossed 3 times. Let $X$ be the sum of the outcomes at each toss. Find the probability of getting a sum of 7 (i.e., $X=7$) at least once out of the three tosses.
My attempt:
The sample space is $6^3=216.$ There are 15 combinations that gives a sum of $7$.
$$P(X=7)=\frac{15}{216}$$
I am stuck here. If a pair of fair dice is tossed 3 times, should the sample space be 36 or 216? Am I on the right path?
Best Answer
This is Bernouly process. The probability that you get 7 at one toss is $$p={6\over 36}={1\over 6}$$
So not getting 7 in 3 trials is $$P' = q^3 = (1-p)^3 = {125\over 216}$$
So the finaly answer is $$ P = 1-P' ={91\over 216}$$