Question

X and Y are loss random variables, with X discrete and Y continuous. The joint density function of X and Y is f(x, y) = [(x+1)e-y/2] / 12 for x = 0, 1, 2 and 0 < y < [infinity]. Find the probability that the total loss, X + Y is less than 2.

Answer 1

Answer:

The probability that the total loss, X + Y is less than 2 is P=0.235

Step-by-step explanation:

We know the joint density function:

$f(x,y)=\frac{(x+1)e^{-y/2}}{12}$

To find the probability that (X+Y)<2, we can divide this in two steps.

- When X=0, Y should be less than 2. This is P(X=0,Y<2).

- When X=1, Y should be less than 1. This is P(X=1, Y<1).

We can calculate P(X=0,Y<2) as:

$P(X=0,Y$

We can calculate P(X=1,Y<1) as:

$P(X=1,Y$

Then

$P(X+Y$