# Probability Question involving dices

Two fair dice are thrown. Given that the total score obtained is even, find the probability of throwing a double.

So I got that the sample space is all the possible outcomes and, |S|= 21.

A = Event that total score obtained is even. |A| = 12

B = Event that double is thrown , |B| = 6

How should i continue thereafter?

I took $\frac{\frac{6}{21}}{\frac{12}{21}}=\frac{1}{2}$

but the given answer is $\frac{1}{3}$

Am i approaching the qn the right way?

As you realised, there are $36$ outcomes, $\{(1,1), (1,2),\ldots,(1,6),(2,1),(2,2),\ldots,(2,6), \ldots, (6,1),\ldots,(6,6)\}$ (of the form first die-result, second die-result).
Of these, half ($18$) have even sum, and of these $6$ have a double. So indeed the right answer should be $\frac{6}{18} = \frac{1}{3}$.