# Q.10.     Find the mean number of heads in three tosses of a fair coin.

S seema garhwal

Let X be success of getting head.

When 3  coins are  tossed  then sample space $=\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}$

$\therefore$  X can be 0,1,2,3

$P(X=0)=P(TTT)=\frac{1}{8}$

$P(X=1)=P(HTT)+P(TTH)+P(HTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=2)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$

$P(X=3)=P(HHH)=\frac{1}{8}$

The probability distribution is as

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

$=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$
$= \frac{3}{8}+ \frac{6}{8}+ \frac{3}{8}$
$=\frac{12}{8}$
$=\frac{3}{2}=1.5$