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"If you roll a fair coin 10 times what is the expected product of number of heads and number of tails?"

Someone answered 25 at at glassdoor.com. My answer would be:

E(k(10-k)) where k is the rv representing the number of heads thrown.

= 10E(k) - E(k^2)

= 10*mean - (var + mean^2)

where mean = np = 10*0.5 = 5, and var = npq = 10*.5*.5 = 2.5 (these are the formulas for the binomial distribution), thus,

= 10*5 - (2.5 + 25) = 22.5

Who is correct?

Thanks.

Nick.

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# Binomial Distribution

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