Question

An unbiased coin is tossed $4$ times. Find the mean and variance of number of heads obtained.

Answer 1

Let $X$ denote the number of heads in the four tosses of the coin.

Then $X$ is a random variable that can take the values $0$,$1,2,3$ or $4$

$P(X=0)$ = Probability of getting no head $\left(TTTT\right)=\frac{1}{16}$

$P(X=1)$ = Probability of getting one head $(HTTT,THTT,TTHT,TTTH)=4\times \frac{1}{16}=\frac{1}{4}$

$P(X=2)$ = Probability of getting two heads $(HHTT,HTHT,HTTH,THHT,THTH,TTHH)=6\times \frac{1}{16}=\frac{3}{8}$

$P(X=3)$ = Probability of getting three head $(HHHT,HHTH,HTHH,THHH)=4\times \frac{1}{16}=\frac{1}{4}$

$P(x=4)$ = Probability of getting four heads $\left(HHHH\right)=\frac{1}{16}$

the probability distribution of $X$:-

$\Rightarrow \sum P_i{x}_i=2$ and

$\Rightarrow \sum P_i{x}_i^2=5$

$\because Mean=\sum P_i{X}_i=2$

Variance $\sum P_i{X}_i^2-{(\sum P_i{X}_i)}^2$

$=5-(2{)}^2=5-4=1$

$Mean=2$ and $Variance=1$