Question:
Consider a die thrown $4$ times. Let $X$ be the number $1$'s and $Y$ be the number $2$'s obtained in the $4$ throws. What is the probability to obtain one number $2$ in the $4$ tries given that the number of $1$'s is odd.
My solution:
Let $A$ be the event that $X$ is $1$ or $3.$ $$P(A) = \binom{4}{1}\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^{3} + \binom{4}{3}\big(\frac{1}{6}\big)^{3}\big(\frac{5}{6}\big) = \frac{65}{162}.$$
Now, let $B$ the event that $Y = 1.$ $$P(A \cap B) = \binom{4}{1, 1, 2}\big(\frac{1}{6}\big)^{2}\big(\frac{2}{3}\big)^{2} + \binom{4}{1 ,3}\big(\frac{1}{6}\big)^4 = \frac{49}{324}.$$
Then the probability of $B$ given $A$ is $ \frac{49}{130} = 0.3769$
Is that the correct answer?