The variance of the number of heads in two tosses of a coin is :

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 = $E(X^2)-E^2(X)$

$\frac{3}{2}-1=\frac{1}{2}$