Binomial Distribution

WCMU101

Question is:

"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.

haruspex

You are right. Another way:
[itex]\sum[/itex]r(n-r).[itex]^{n}[/itex]C[itex]_{r}[/itex] =
[itex]\sum[/itex]n!/(r-1)!/(n-r-1)! =
[itex]\sum[/itex]n(n-1)(n-2)!/(r-1)!/(n-r-1)! =
n(n-1)[itex]\sum[/itex][itex]^{n-2}[/itex]C[itex]_{r-1}[/itex] =
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