Skip to content

CA Foundation · Quantitative Aptitude · Theoretical Distributions

A fair coin is tossed twice. If X denotes the number of heads, what are E(X) and Var(X)?

E(X) is 1 and Var(X) is 0.5. The probabilities of 0, 1 and 2 heads are 0.25, 0.5 and 0.25, giving a mean of 1 and E(X squared) of 1.5, so variance is 0.5. This agrees with binomial results np and npq.

  1. AE(X) = 1, Var(X) = 1
  2. BE(X) = 0.5, Var(X) = 0.5
  3. CE(X) = 1, Var(X) = 0.5Correct
  4. DE(X) = 2, Var(X) = 0.5

Explanation

P(X=0)=0.25, P(X=1)=0.5, P(X=2)=0.25. E(X) = 0.5 + 0.5 = 1. E(X^2) = 0.5 + 1 = 1.5, so Var = 1.5 - 1 = 0.5. This matches binomial np = 1 and npq = 0.5. Variance 1 wrongly equals the mean.

Did you get it right without looking?

One question tells you little. A timed set on Theoretical Distributions shows your real accuracy, how long you take and where you lose marks.

Practise Theoretical DistributionsFree diagnostic mock

More Theoretical Distributions questions