Question

Let X be a binomial(5, 0.2) random variable. Let Y be a discrete random variable that is independent of X, such that Y = 1 with probability 0.2 and Y = 0 with probability 0.8. What is the probability that the sum of X and Y is less than or equal to 3?

Answer 1

$\begin{aligned} X & \sim \operatorname{Bin}(5,0.2) \Rightarrow P(X=x)=\left(\begin{array}{l} 5 \\ x \end{array}\right)(0.2)^{x}(1-0.2)^{5-x} \\ \end{aligned}$

$\begin{aligned} & P(Y=0)=0.8, \quad P(Y=1)=0.2 \\ \therefore & P(X+Y \leqslant 3)=P(Y=0, X \leqslant 3)+P(Y=1, X \leqslant 2) \\ =& 0.8 P(X \leqslant 3)+0.2 P(X \leqslant 2) \\ =& 0.8(0.32768+0.4096+0.2048+0.0512) +0.2(0.32768+0.4096+0.2048) \\ =& 0.98304 \end{aligned}$