# Probability of at least once out of the three tosses

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?

• You have to show how you worked out $\frac{15}{216}$. The method you used could have a problem , or it could be a calculation : either way, showing it would make life easier for somebody trying to answer your question. Did you enumerate the possibilities, or notice any independence etc.? – астон вілла олоф мэллбэрг Oct 11 '18 at 13:30
• Hint: find the probability of the complement of that event, i.e. the probability that a sum $7$ does not arise at the $3$ tosses. – drhab Oct 11 '18 at 13:32
• Your result $15/216$ accords with the probability that the sum of three dice rolled once (or one die rolled three times) will be $7$. – Barry Cipra Oct 11 '18 at 13:48

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 $$P = 1-P' ={91\over 216}$$