Question

In tossing a fair coin twice, find the probability of getting at least one head

Answer 1

When a fair coin is tossed twice, $S=\{HH,TT,HT,TH\}\therefore n\left(S\right)=4$
Let $B$ be the event of getting at least one heads
$B=\{HH,HT,TH\}\therefore n\left(B\right)=3$
$\therefore P\left(B\right)=\frac{n\left(B\right)}{n\left(S\right)}=\frac{3}{4}$