Question

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

• A. $\frac{1}{6}$
• B. $\frac{3}{4}$
• C. $\frac{2}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

In tossing a coin twice, the sample space, $S=HH,HT,TH,TT$
Therefore, $n\left(S\right)=4$
Let $C$ be the event of getting exactly one head.
$C=HT,TH$

Therefore, $n\left(C\right)=2$
$P\left(C\right)=$ $\frac{n\left(A\right)}{n\left(S\right)}=\frac{2}{4}=\frac{1}{2}$