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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?

Thanks in advanced!

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  • $\begingroup$ OH realised my mistake. it should be 36 outcomes instead of 21 $\endgroup$ – qwerty23 Nov 16 '14 at 12:38
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    $\begingroup$ If your question is not a question anymore (you realized your mistake) then make use of the possibility to delete your question. $\endgroup$ – drhab Nov 16 '14 at 13:09
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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 $\frac{6}{18} = \frac{1}{3}$.

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