From Ostaszewski, 2007:
An insurance policy covers two employees of a company. The policy will reimburse no more than one loss per employee per year. It reimburses the full amount of the loss up to a company wide maximum of $8000$. The probability of an employee incurring a loss is $40$%, and is independent of the other employee's losses. The amount of each loss is uniformly distributed on $(1000, 5000)$. Given that one of the employees has incurred a loss in excess of $2000$, determine the probability that losses will exceed reimbursements.
The way to solve this is to make a square with vertices $(1000,1000), (5000,1000), (5000,5000),$ and $(1000,5000)$.
Then, to find the condition that one of the employees incurred $\text {loss} >2000$, one must find the area that goes from $2000$ to $5000$ on one of the axes, and from $1000$ to $5000$ on the other, for a total area of $3000\cdot 4000$ in the denominator.
The probability that losses exceed reimbursement is $P(X+Y>8000)$, so the area is the triangle with vertices $(3000,5000), (5000,5000),$ and $(5000,3000)$, and so the numerator will be ${1\over2}\cdot 2000^2$.
What is confusing me is how to account for fact that the probability of an employee incurring a loss is $40$%. I would have thought that the total probability of at least one employee incurring a loss is $.40\cdot.60 +.60\cdot.40 + .40\cdot.40$, and the probability that both employees incur a loss (which is necessary for losses to exceed reimbursements) is $.40 \cdot .40$, and therefore the solution should be
$$.40 \cdot .40\cdot {1\over2} \cdot 2000^2 \over (.40\cdot.60 +.60\cdot.40 + .40\cdot.40)\cdot 3000 \cdot 4000$$
However the solution says to just do:
$$.40\cdot {1\over2} \cdot 2000^2 \over 3000 \cdot 4000$$
What is wrong with how I wanted to do it?
Best Answer
I think there is some ambiguity in the question. I see two interpretations:
Interpretation (1): (This is how I would have interpreted the question.)
The probability that at least one employee incurred a loss in excess of $2,000$ is $P[L_1 > 2000 \text{ or } L_2 > 2000] = 1-P[L_1 \le 2000 \text{ and } L_2 \le 2000] =1-(0.6+{2000-1000 \over 5000-1000}0.4)^2$.
The probability that the loss exceeds $8,000$ can be computed graphically as $P[L_1+L_2 > 8000] ={1 \over 2} P[L_1 > 3000]^2={1 \over 2} ({5000-3000 \over 5000-1000}0.4)^2$.
Note that the second event implies the first. Hence $P[L_1+L_2 > 8000 | L_1 > 2000 \text{ or } L_2 > 2000] = {P[L_1+L_2 > 8000 ] \over P[L_1 > 2000 \text{ or } L_2 > 2000] } = {2 \over 51}$.
Interpretation (2): (This matches your solution above.)
We have $P[L_1 > 2000] = 1-(0.6+{2000-1000 \over 5000-1000}0.4)$. Again, this event implies the first so $P[L_1+L_2 > 8000 | L_1 > 2000] = {P[L_1+L_2 > 8000 ] \over P[L_1 > 2000 ]} = {1 \over 15}$.