Practicing Success
The variance of the number of heads in two tosses of a coin is : |
$\frac{1}{2}$ 1 $\frac{1}{8}$ $\frac{3}{8}$ |
$\frac{1}{2}$ |
The correct answer is Option (1) → $\frac{1}{2}$ No of tosses = 2 $P(X=0)={^2C}_0(\frac{1}{2})^2=\frac{1}{4}$ $P(X=1)={^2C}_1(\frac{1}{2})^2=\frac{1}{2}$ $P(X=2)={^2C}_2(\frac{1}{2})^2=\frac{1}{4}$ $E(X)=∑P_ix_i=0×\frac{1}{4}+1×\frac{1}{2}+2×\frac{1}{4}=1$ $E(X^2)=∑P_ix_i^2=0^2×\frac{1}{4}+1^2×\frac{1}{2}+2^2×\frac{1}{4}=\frac{3}{2}$ Variance = $E(X^2)-E^2(X)$ $\frac{3}{2}-1=\frac{1}{2}$ |