Find $\frac{P(X+Y\leq 2)}{P(X+Y\geq 5)}$

Question

Let $X$ follow $B(5, 1/2)$ and $Y$ follow $U(0, 1)$. Then find

$$\frac{P(X+Y\leq 2)}{P(X+Y\geq 5)}.$$

How to do that? I am confused sum of discrete and continuous random variable. pleas help.

Answer 1

Basically $Y$ rounds $X$ up a little bit. It will never be exactly $0$ or $1$, so $P(X+Y) \le 2$ is the same as $P(X) \le 1$ and similarly for the denominator.

Answer 2

Since $X \sim {\cal B}(5,1/2)$ and $Y \sim U(0,1)$, we can calculate $P(X+Y \le 2)$ and $P(X+Y \ge 5)$ by taking discrete sum on $P(X = k)$, where $k \in \{0,\dots,5\}$.

$\require{cancel}$ \begin{align} P(X+Y\le2) &= \sum_{k=0}^5 P(Y \le 2-k) P(X = k) \\ &= \cancelto{1}{P(Y \le 2)} P(X = 0) + \cancelto{1}{P(Y \le 1)} P(X = 1) \\ &+ \cancelto{0}{P(Y \le 0) P(X = 2)} + \cdots \\ &= P(X = 0) + P(X = 1) \\ &= \frac{1+5}{32} \end{align}

Similarly,

\begin{align} P(X+Y\ge5) &= \sum_{k=0}^5 P(Y \ge 5-k) P(X = k) \\ &= P(Y \ge 1) P(X = 4) + P(Y \ge 0) P(X = 5) \\ &= P(X = 5) \\ &= \frac{1}{32} \end{align}

Hence $$\frac{P(X+Y\leq 2)}{P(X+Y\geq 5)} = \frac{6/32}{1/32} = 6.$$