Question

In tossing a fair coin twice, find the probability of getting no heads

Answer 1

When a fair coin is tossed twice, $S=\{HH,TT,HT,TH\}\therefore n\left(S\right)=4$
Let $C$ be the event of getting no heads
$C=\{T,T\}\therefore n\left(C\right)=1$
$\therefore P\left(C\right)=\frac{n\left(C\right)}{n\left(S\right)}=\frac{1}{4}$