- #1
WCMU101
- 14
- 0
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.
"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.