A computer store has purchased three computers of a certain type at $500 apiece. It will sell them for $1000 apiece. The manufacturer has agreed to repurchase any computers still unsold after a specified period at $200 apiece. Let X denote the number of computers sold, and suppose that p(0) = 0.1, p(1) = 0.2, p(2) = 0.3, p(3) = 0.4 With h(X) denoting the profit associated with selling X units, the given information implies that h(X) = revenue−cost = 1000X+200(3-X) - 1500 = 800X-900. What is the expected profit?

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Answer:

The expected profit will be of 700 dollars

Explanation:

The expected profit will be the multiplication of the outcomes by their probability:

We have 10% chance of selling zero thus:

1,000(0) + 200(3-0) - 1,500 = -900

We have 20% chance of selling one thus:

1,000(1) + 200(3-1) - 1,500 = -100

We have 30% chance of selling two thus:

1,000(2) + 200(3-2) - 1,500 = 700

We have 40% chance of selling all thus:

1,000(3) + 200(3-3) - 1,500 = 1,500

[tex]\left[\begin{array}{ccc}weight&outcome&w.profit\\0.1&-900&-90\\0.2&-100&-20\\0.3&700&210\\0.4&1,500&600\end{array}\right][/tex]

We add each weighted profit to get the expected result:

-90 - 20 + 210 + 600 = 700

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