Question

Find the mean number of heads in three tosses of fair coin.

Answer 1

Let X denote the success of getting heads. 

Therefore, the sample space is 

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} 

It can be seen that X can take the value of 0, 1, 2, or 3.

P(X=0)=P(TTT)

=P(T).P(T).P(T)

=\(\frac{1}{2}X\frac{1}{2}X\frac{1}{2}\)=\(\frac{1}{8}\)

∴ P (X = 1) = P (HHT) + P (HTH) + P (THH)

\(\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}\)=\(\frac{3}{8}\)

∴P(X = 2) = P (HHT) + P (HTH) + P (THH)

\(\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}X\frac{1}{2}\)=\(\frac{3}{8}\)

∴P(X=3)=P(HHH)=\(\frac{1}{2}X\frac{1}{2}X\frac{1}{2}\)

=\(\frac{1}{8}\)

Therefore, the required probability distribution is as follows.

X	0	1	2	3
P(X)	\(\frac{1}{8}\)	\(\frac{3}{8}\)	\(\frac{3}{8}\)	\(\frac{1}{8}\)

Mean of XX(X),μ=∑X_iP(X_i)

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

=\(\frac{3}{8}+\frac{3}{8}+\frac{3}{8}\)

=\(\frac{3}{2}=1.5\)