# What is the probability of rolling a number greater than or equal to 8 with the sum of two dice, given that at least one of the dice must show a 6?

$\frac{9}{36} = \frac{1}{4}$

#### Explanation:

I'm going to solve this in this way - we know that there are $6 \cdot 6 = 36$ different results we can get from rolling 2 dice, so in this probability fraction, 36 is the denominator.

We're constrained that one die must be a 6 and the total must be equal to or greater than 8.

When die A is 6, die B can be 2-6 but can't be 1, so we have 5 acceptable results.

When die B is 6, die A can be 2-6 but can't be 1, so we have 5 more acceptable results.

Now note that we've just double counted the (6,6) result, so we take away 1 good result, and end up with 9 and a total probability of

$\frac{9}{36} = \frac{1}{4}$