## Binomial Distribution

Question is:

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

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.
 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Recognitions: Homework Help Science Advisor You are right. Another way: $\sum$r(n-r).$^{n}$C$_{r}$ = $\sum$n!/(r-1)!/(n-r-1)! = $\sum$n(n-1)(n-2)!/(r-1)!/(n-r-1)! = n(n-1)$\sum$$^{n-2}$C$_{r-1}$ = n(n-1).2^(n-2) Dividing by 2^n to get the average: n(n-1)/4 For n = 10 this gives 22.5