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October 2nd, 2007, 04:59 PM   #1
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The expected gain, given a Binomial Distribution?

I understand the concept of a Binomial Distribution, and mean and variance. But asked to calculate expected gain I'm stumped!
Here's a problem:

10 motors are packaged for sale in a warehouse. The motors sell for 100$ each, but a double-your-money-back guarantee is in effect for any defectives the purchaser may receive. Find the expected net gain for a seller if the probability of any one motor being defective is 0.08

Would it be the 10 (probability of an undefective motor times 100 + probability of a defective motor times 200?)


Help, please!!!!

The answer is 840$!
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October 2nd, 2007, 05:44 PM   #2
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For each of the 10 motors:

The probability of it being undefective is 1-0.08 = 0.92,
in this case the seller receives $100.

The probability of it being defective is 0.08,
in this case the seller receives $100 and then pays back to the customer double 100 = $200, which gives a net outflow of $100 for the seller.

Hence the only part you need to change in your answer is the (-200) which should be (-100) instead.

10 ( Pr{undefective}*100 + Pr{defective}*(-100) )
= 10 ( 0.92 * 100 + 0.08 * (-100) )
= $840.
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