Question

Find the mean number of heads in three toses of a fair coin.

Answer 1

Probability of getting a head and tail in a single toss of a coin $P\left(H\right)=P\left(T\right)=\frac{1}{2}$

The sample space of three toses of a coin is

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

It can be seen that random variable X(say) can take the value of 0,1,2 ot 3.

P(X=0)=P(probability of getting all tails)=P{TTT}=$\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

P(X=1)=P (probability of getting one head and two tails)

=P{TTH}+P{THT}+P{HTT}

$=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

P(X=2)=P(Probability of getting two heads and one tail)

=P{HHT}+P{HTH}+P{THH}

=$=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

P(X=3)=P(probability of getting three heads) =$PHHH=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

Therefore, the required probability distribution is as follows:

$\begin{array}{ccccc}X& 0& 1& 2& 3\\ P\left(X\right)& \frac{1}{8}& \frac{3}{8}& \frac{3}{8}& \frac{1}{8}\end{array}$

Hence, mean =$\sum XP\left(X\right)=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$

$=\frac{3}{8}+\frac{6}{8}+\frac{3}{8}=\frac{12}{8}=\frac{3}{2}=1.5$