Two fair dice are rolled. What is the conditional probability that at least one of the dice lands on $5$ given that their sum is $11$?
$\textbf{I've solved it as following and it is scored as wrong answer:}$
$P(E|F) = \frac{P(E \cap F)}{P(F)}$
$ Event F = \left \{(1,5), (2,5),(3,5),(4,5),(5,5),(6,5), (5,1),(5,2),(5,3),(5,4),(5,6)\right \} $
$P(F) = \frac{11}{36}$
$ Event E = \left \{ (5,6), (6,5) \right \}$
$ E \cap F = \left \{(5,6), (6,5)\right \}$
$ P(E \cap F) = \frac{2}{36}$
$ P(E|F) = \large\frac{\frac{2}{36}}{\frac{11}{36}} = 0.1818$
How to solve this problem correctly?
You have switched up $E$ and $F$. Looking at your solution, $E$ seems to be the event that the two dice sum to $11$, while $F$ is the event that at least one of the two dice rolls a $5$. You computed $P(E|F)$, which is the probability that the sum of the two dice is $11$, given that at least one of the two dice rolls a $5$. If you notice, this is actually swapped of what you are trying to find, which is the probability that at least one of the two dice rolls a $5$, given that the sum of the two dice is $11$.
Try computing $P(F|E)$ instead; you will arrive at the correct answer.