Question

Find the probability distribution of number of heads in two tosses of a coin.

Answer 1

Given that a single coin is tossed twice, therefore the sample space $S=\{HH,HT,TH,TT\}$

Since the $4$ events are equally likely, we have

$P\left(HH\right)=P\left(HT\right)=P\left(TH\right)=P\left(TT\right)=\frac{1}{4}$

Let $X$ be the number of heads in $S$ such that $X\left(HH\right)=2,X\left(HT\right)=1,X\left(TH\right)=1,X\left(TT\right)=0$ , It follows that

$P(X=0)=P\left(TT\right)=\frac{1}{4}$

$P(X=1)=P(TH,HT)=\frac{1}{2}$

$P(X=2)=P\left(HH\right)=\frac{1}{4}$

Therefore the probability distribution table is shown below:

$X$ $0$ $1$ $2$

$P\left(X\right)$ $\frac{1}{4}$ $\frac{1}{2}$ $\frac{1}{4}$