EX 5.1

• Anyone purchasing an insurance policy for a home or automobile:
• Must specify a deductible amount.
  • Deductible: the amount of loss the policyholder absorbs before the insurance company begins paying out.
• Suppose a particular company offers:
  • Auto deductibles: \100$,$$500$,$$1000$.
  • Homeowner deductibles: \500$,$$1000$,$$2000$.
• Consider randomly selecting someone with both auto and homeowner insurance from this company.
• Let:
  • $X$: the auto policy deductible amount.
  • $Y$: the homeowner policy deductible amount.
• The joint pmf of $X$ and $Y$ is represented in a joint probability table.

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

• According to the joint pmf:
  • There are nine possible $\left(X,Y\right)$ pairs:
  • Examples: $\left(100,500\right)$, $\left(100,1000\right)$, …, $\left(1000,5000\right)$.
• The probability of $\left(100,500\right)$ is:
  • $p\left(100,500\right)=P\left(X=100,Y=500\right)=0.30$.
• General properties of the pmf:
  • $p\left(x,y\right)\ge 0$ for all $\left(x,y\right)$ pairs.
  • The sum of the nine displayed probabilities equals 1.
• To compute the probability $P\left(X=Y\right)$:
  • Sum $p\left(x,y\right)$ over the pairs where the deductible amounts are identical:
  • This includes two $\left(x,y\right)$ pairs.

$$\begin{array}{rl}P(X=Y)& =p(500,500)+p(1000,1000)\\ & =0.15+0.10\\ & =0.25\end{array}$$

• To find the probability that the auto deductible amount is at least $500:
• Calculate the sum of all probabilities for $\left(x,y\right)$ pairs where:
  • $x\ge 500$.
• This corresponds to:
  • The sum of probabilities in the bottom two rows of the joint probability table.

$$\begin{array}{rl}P(X\ge 500)& =0.15+0.20+0.05+0.10+0.10+0.05\\ & =0.65\end{array}$$