Three distinct fair dice are thrown. The probability that $4$ appears on two dice given that $5$ occurs on atleast one dice.
I counted the number of total cases when atleast one five occurs. This can be counted as (exactly 1+ exactly 2+exactly 3) = $ 75+15+1$. The favourable cases are $3$. So probability should be $3/91$ according to me. However, the answer that is given in the source from where the question has been taken is something else. So, I just wanted to confirm whether my method is correct.
$A$ is event of $4$ appearing exactly on two of the three fair dice.
$B$ is the event of $5$ appearing on at least one dice.
$P(A \cap B) = {3 \choose 1} \frac{1}{6^3}$ (two $4$ and one $5$)
$P(B) = 1 - (\frac{5}{6})^3$ (this is probability of at least one $5$ appearing).
Desired conditional probability $P(A|B) = \displaystyle \frac{P(A \cap B)}{P(B)} = \frac{{3 \choose 1} \frac{1}{6^3}}{1 - (\frac{5}{6})^3} = \frac{3}{91}$.