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:

$P(X=0,Y<2)=\int_0^2f(0,2)dy=\int_0^2 \frac{e^{-y/2}}{12} dy\\\\P(X=0,Y<2)=-(e^{-2/2}-e^{-0/2})/6=-(0.37-1.00)/6\\\\P(X=0,Y<2)=-(-0.63)/6=0.105$

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

$P(X=1,Y<1)=\int_0^1f(1,1)dy=\int_0^1 \frac{2e^{-y/2}}{12} dy\\\\P(X=1,Y<1)=-(e^{-1/2}-e^{-0/2})/3=-(0.61-1.00)/3\\\\P(X=1,Y<1)=-(-0.39)/3=0.130$

Then

$P(X+Y<2)=P(X=0,Y<2)+P(X=1,Y<1)\\\\P(X+Y<2)=0.105+0.130=0.235$