An unbiased die is thrown $3$ times. Given that the first throw is a $5$, find the conditional probability of getting $16$ as a sum.
Here's what I have done. $$P(\text{Sum}=16\mid\text{First throw}=5)=\frac{P(\text{Sum}=16\text{ AND First throw}=5)}{P(\text{First throw}=5)},$$
$$P(\text{First throw}=5)=\frac{5\times6\times6}{6\times6\times6}.$$
Now, I have a confusion here in the following regarding choosing the sample space. To count $$P(\text{Sum}=16\text{ AND First throw}=5),$$ the number of elementary events that are favorable is $2$, and they are $5+5+6$, $5+6+5$: since the first throw is already $5$, there are $2$ choices to make the remaining $11$ so as to make the sum equal to $16$. But what is the sample space? Is it $5\times6\times6$, the number of ways to get as the first throw a $5$ and the remaining two throws out of $\{1,2,3,4,5,6\}$?
We know that the first roll is $5$. Thus the second and third die throws must sum to $11$ (because $16 - 5 = 11$).
With six sided dice, there are only two possible combinations for two throws to sum to $11$; that is, the first throw being $5$ and the second throw being $6$, or the first throw being $6$ and the second throw being $5$.
There are $36$ possible combinations and $2$ are valid, meaning the probability will be $\dfrac{2}{36} = \dfrac{1}{18}$.