EXAMPLE 5.1

Anyone who purchases an insurance policy for a home or automobile must specify a deductible amount, the amount of loss to be absorbed by the policyholder before the insurance company begins paying out. Suppose that a particular company offers auto deductible amounts of \ {100},$ {500}$,and$$ {1000}$,andhomeownerdeductibleamountsof$$ {500},$ {1000}$,and$$ {2000}$. Consider randomly selecting someone who has both auto and homeowner insurance with this company, and let

• $X=$ the amount of the auto policy deductible
• $Y=$ the amount of the homeowner policy deductible.

The joint pmf of these two variables appears in the accompanying joint probability table:

$p(x,y)$	500	1000	5000
100	.30	.05	0
500	.15	.20	.05
1000	.10	.10	.05

According to this joint pmf, there are nine possible $\left(X,Y\right)$ pairs: $\left(100,500\right),(100$, $1000),\dots$, and finally $\left(1000,5000\right)$. The probability of $\left(100,500\right)$ is $p\left(100,500\right)=P\left(X=100,Y=500\right)=.30.$ Clearly $p\left(x,y\right)\ge 0$, and it is easily confirmed that the sum of the nine displayed probabilities is 1.

The probability $P\left(X=Y\right)$ is computed by summing $p\left(x,y\right)$ over the two $\left(x,y\right)$ pairs for which the two deductible amounts are identical:

$$P\left(X=Y\right)=p\left(500,500\right)+p\left(1000,1000\right)=.15+.10=.25$$

Similarly, the probability that the auto deductible amount is at least $500 is the sum of all probabilities corresponding to $\left(x,y\right)$ pairs for which $x\ge 500$; this is the sum of the probabilities in the bottom two rows of the joint probability table:

$$P\left(X\ge 500\right)=.15+.20+.05+.10+.10+.05=.65$$