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Two fair six-sided dice are rolled. If the sum of the numbers obtained is $4$, find the probability that the numbers obtained on both dice are even.

I'm not sure if it's $\frac{1}{36}$ or $\frac{1}{3}$.

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  • $\begingroup$ Number of states, sum of two even is four : $$\#\{(2,2)\}=1$$ Number of states, sum of two dice is four : $$\#\{(1,3),(2,2),(3,1)\}=3$$ $\endgroup$ – Fardad Pouran Jun 15 '14 at 5:03
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$4$ can obtained by $1+3,2+2,3+1$

So, the probability $\displaystyle=\frac{\text{No. of favorable cases}}{\text{No. of possible cases}}=\frac13$

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  • $\begingroup$ That is because the sum has already been given and the size of our sample space reduces to 3? $\endgroup$ – Pratik Jun 15 '14 at 5:04
  • $\begingroup$ @Pratik, Exactly $\endgroup$ – lab bhattacharjee Jun 15 '14 at 5:04
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The important point is that you are first given that the sum is $4$.

This can only happen rolling $(1,3), (2,2)$ or $(3,1)$.

So given that the sum is $4$, we now have a sample size of $3$, out of which only $1$ entry, i.e., $(2,2)$, has both dice showing an even number.

Therefore, the answer is $1/3$.


If, however, you were just asked: What is the chance of rolling $(2,2)$?

Then your sample size would be all the possible rolls with two dice, and you would see that only $1$ of the $36$ possibilities is $(2,2)$. In this case, different from yours, the answer would be $1/36$.

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    $\begingroup$ Thank you. That was very helpful $\endgroup$ – Pratik Jun 15 '14 at 5:11
  • $\begingroup$ @Pratik You're welcome! Hopefully the problem is clearer now... $\endgroup$ – Benjamin Dickman Jun 23 '14 at 7:23
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This problem is related to conditional probability or Bayes theorem. The probability P(A) of getting the sum 4 is 3/36. Probability of getting even numbers on each dice and sum 4 is $P(A \cap B)= $1/36. So conditional probability is

$P(A \cap B)/P(A) =\frac{1/36}{3/36}=1/3$

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