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An insurance company writes a policy to the effect that an amount of money $A$ must be paid if some event $E$ occurs within a year. If the company estimates that $E$ will occur within a year with probability $p$, what should it charge the customer so that its expected profit will be $10$ percent of $A$?

How can I set up this problem?

Assuming the random variable is the amount of profit, it seems like I need information on the probability that profit = $\$0$ and probability that profit negative. Is the question ambiguous as to how many times the company will pay for the event?

Otherwise, I have the equation $E[X]=-Ap+A(1-p)=\frac{A}{10}$, with which I find $p=\frac{9}{20}$ which is of no use.

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Let $C$ be the amount the company charges. Then the expected profit is equal to $C-pA$ (income minus expected expense), or if you prefer $(C-A)(p) +(C)(1-p)$. The second version is because with probability $p$, the company's net income will be $C-A$, and with probability $1-p$ the net income will be $C$.

We want expected net income to be equal to $\dfrac{A}{10}$. So set $C-pA=\dfrac{1}{10}A$ and solve for $C$. We get $$C=A\left(p+\frac{1}{10}\right).$$

As to your question, yes, there is some imprecision in the question. But we are expected to assume that the payout will be given out at most once. Think for example in terms of life insurance.

The number $p$ is a parameter of the problem, presumably known more or less accurately by the insurance company.

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