If we are given that $Y=3$, find the probability that $Y$ is the sum of 2 dice. A fair die is tossed. If the outcome is $n$ then $n$ fair dice are tossed and $Y$ is the sum of these $n$ dice. If we are given that $Y=3$, find the probability that $Y$ is the sum of 2 dice.
 A: Let $A$ be the event that the first toss resulted in a $2$, and let $B$ be the event the second sum was $3$. We want $\Pr(A|B)$. Note that
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
We need the two probabilities on the right, and start with the harder one, $\Pr(B)$.
The result on tossing the die or dice the second time can be $3$ in three ways: (i) We got a $1$ on the first die, and then a "sum" of $3$ the next time; (ii) We got a $2$ on the first toss, and a sum of $3$ using $2$ dice the next time; (iii) We got a $3$ on the first die, and then a sum of $3$ from $3$ dice. 
We calculate the probabilities of these and add up.  (i) The probability of this is the probability of a $1$ followed by a $1$. This is $\frac{1}{6}\cdot\frac{1}{6}$;  (ii) With probability $\frac{1}{6}$ we got an initial toss of $2$. The probability that this is followed by a sum of $3$ is $\frac{2}{36}$, for a probability of $\frac{1}{6}\cdot\frac{2}{36}$; (iii) This is also straightforward.
Now find $\Pr(A\cap B)$ and finish the calculation. 
