Question

A coin is tossed $3$ times. The probability of obtaining two heads will be

• A. $3/8$
• B. $1/2$
• C. $1$
• D. $2$

Answer 1

The correct option is A $3/8$

$N\left(S\right)=2^3=8$
$E=\{HHT,HTH,THH\}\therefore n\left(E\right)=3$
Hence required probability $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{8}$