The mean of the number of heads in a simultaneous toss of three coins is:

$\frac{3}{2}$

The correct answer is Option (3) - $\frac{3}{2}$

$P(H)=\frac{1}{2}$

$P(T)=\frac{1}{2}$

No of tosses = 3

so $P(X=0)={^3C}_0(\frac{1}{2})^0(\frac{1}{2})^3=\frac{1}{8}$

$P(X=1)={^3C}_1(\frac{1}{2})(\frac{1}{2})^2=\frac{3}{8}$

$P(X=2)={^3C}_2(\frac{1}{2})^2(\frac{1}{2})=\frac{3}{8}$

$P(X=3)={^3C}_3(\frac{1}{2})^3(\frac{1}{2})^0=\frac{1}{8}$

so mean = $∑x_ip_i$

$=0×\frac{1}{8}+1×\frac{3}{8}+2×\frac{3}{8}+3×\frac{1}{8}$

$=\frac{3+6+3}{8}=\frac{3}{2}$