# If you roll a pair of dice, what is the probability of rolling either a single 5 or a sum that is an even number?

Oct 28, 2015

$\frac{2}{3}$

#### Explanation:

Note that total number of possible cases are ${6}^{2} = 36$

Getting a single $5$ ( Say event $A$ ) means situation like as $\left(1 , 5\right) , \left(2 , 5\right) , \left(3 , 5\right) , \left(4 , 5\right) , \left(6 , 5\right) , \left(5 , 1\right) , \left(5 , 2\right) , \left(5 , 3\right) , \left(5 , 4\right) , \left(5 , 6\right)$ i.e. $10$ cases and we have $18$ cases when sum is an even number ( Say event $B$ ) . But these two events are not mutually exclusive. Here $\left(1 , 5\right) , \left(3 , 5\right) , \left(5 , 1\right) , \left(5 , 3\right)$ i.e. $4$ cases where we get a single $5$ as well as sum is a even number ( Say event $A \cap B$ ) .

So we have number of favorable cases to our event $= n \left(A\right) + n \left(B\right) - n \left(A \cap B\right) = 10 + 18 - 4 = 24$

So required probability $= \frac{24}{36} = \frac{2}{3}$
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