# What is the probability of rolling at least one $7$, $11$, or doubles in an experiment consisting of two rolls?

The experiment is rolling a pair of dice, two times. Rolling a $7$, $11$, or doubles is a success, rolling anything else is a failure.

What is the probability of succeeding on at least one of the two rolls.

I calculate the probability of succeeding in a one-roll experiment to be $\frac{14}{36}$:

There are six ways to roll a $7$, six ways to roll doubles, and two ways to roll an $11$, out of $36$ possible outcomes.

For two consecutive rolls, I think that the sample space of outcomes would be $36\cdot 36$.

How would you figure out how many outcomes would succeed in a two-roll experiment? What is the general way to figure out things like this?

I agree that in a single roll we have: $P(\mathrm{success})=\frac{14}{36}$. Now, instead of counting outcomes of two rolls, we can calculate the probability of succeding on at least one of the two rolls (assuming independence of the rolls):
$$P(\text{at least one success in two rolls})=1-P(\text{no successes in two rolls})\\ =1-P(\text{no success in a single roll})^2=1-\left(\frac{22}{36}\right)^2.$$
The probability of failure each roll is $1-\mathrm{P}(7)-\mathrm{P}(11)-\mathrm{P}(\text{pair})=1-\frac{6}{36}-\frac{2}{36}-\frac{6}{36}=\frac{11}{18}$. The probability of failing twice is $\left(\frac{11}{18}\right)^2$. Thus the probability of succeeding at least once is $$1-\frac{121}{324}=\frac{203}{324}$$