I am taking P tomorrow and can't quite understand why my answer isn't correct. It is TIA #2 problem 21. It says:
A company insures 2 machines fro maintenance. In a given year, each machine will require maintenance with probability 1/3, and if maintenance is required, the cost will be a uniform random variable between 0 and 4,000. The insurance policy has an annual payment limit of 6,000 for both machines combined, what is the expected annual payment?
So my logic was pretty simple. The probability that neither machine fails is 4/9, the probability that one machine fails is 4/9, and the probability that both fail is 1/9.
If neither fail, E[Y] = 0. If one fails, E[Y] = 2,000. If both fail, the cost becomes a uniform random variable between 0 and 8,000, so I integrate from zero to the policy limit, of the survival function of f(y), and get 3750.
So E[Y] = (4/9)(0)+(4/9)(2000)+(1/9)(3750) = 1305.
But my answer is wrong.
A company insures 2 machines fro maintenance. In a given year, each machine will require maintenance with probability 1/3, and if maintenance is required, the cost will be a uniform random variable between 0 and 4,000. The insurance policy has an annual payment limit of 6,000 for both machines combined, what is the expected annual payment?
So my logic was pretty simple. The probability that neither machine fails is 4/9, the probability that one machine fails is 4/9, and the probability that both fail is 1/9.
If neither fail, E[Y] = 0. If one fails, E[Y] = 2,000. If both fail, the cost becomes a uniform random variable between 0 and 8,000, so I integrate from zero to the policy limit, of the survival function of f(y), and get 3750.
So E[Y] = (4/9)(0)+(4/9)(2000)+(1/9)(3750) = 1305.
But my answer is wrong.
Help with E[Y] problem for payment
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