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

$P(X=x)=(\begin{matrix} 5 \\ x \end{matrix})0.2^x(1-0.2)^{5-x}$

$P(Y=0)=0.8,P(Y=1)=0.2$

$P(X+Y≤3)=P(Y=0,X≤3)+P(Y=1,X≤2)$

$=0.8P(X≤3)+0.2P(X≤2)$

$=0.8(0.32768+0.4096+0.2048+0.0512)+0.2(0.32768+0.4096+0.2048)$

$=0.98304$