Probability question

The probability distribution of $X$ is given by
$$\mathbb{P}[X = n] = \binom{5}{n} \left(\frac{1}{2}\right)^{5}$$,
as there are $\binom{5}{n}$ ways to choose which cups have First Flush in them, and each way has a probability of $1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2$ of occuring.

The probability distribution of $Y$ is given by
$$P[Y = 0] = 0 \text{ and } \mathbb{P}[Y = n] = \binom{4}{n-1} \cdot \left(\frac{1}{2}\right)^4 \text{ for } 1 \leq n \leq 5 $$
as the number of correct guesses is $1 + (\text{number of cups matching the first})$. The number of ways to choose  which $n-1$ of the $4$ remaining cups agree with the first is $\binom{4}{n-1}$, and each way has a probability of $1/2 \cdot 1/2 \cdot 1/2 \cdot 1/2$ of occurring.

The only outcome in which $X=1$ and $Y=1$ are when the only the first cup is filled with First Flush. The probability of this single outcome is $(1/2)^{5}$.

The probability that $X = 3$ and $Y=1$ is zero. Since either the connoisseur only successfully guesses all cups with First Flush, or only successfully guesses all cups with Second Flush, we must either have $X = Y$ or $X+Y=5$ at each outcome. However, $1 \neq 3$ and $1+3 = 4 \neq 5$.

The variables $X$ and $Y$ are not independent. Note that $\mathbb{P}[X=3, Y=1] = 0$, yet neither $\mathbb{P}[X=3]$ or $\mathbb{P}[Y=1]$ is zero. Hence $\mathbb{P}[X=3, Y=1] \neq \mathbb{P}[X=3]\mathbb{P}[Y=1]$.