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October 2nd, 2007, 04:59 PM  #1 
Newbie Joined: Oct 2007 Posts: 1 Thanks: 0  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 doubleyourmoneyback 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$! 
October 2nd, 2007, 05:44 PM  #2 
Member Joined: Mar 2007 Posts: 57 Thanks: 0 
For each of the 10 motors: The probability of it being undefective is 10.08 = 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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binomial, distribution, expected, gain 
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