Question

A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is

• A. $\frac{1}{8}$
• B. $\frac{1}{2}$
• C. $\frac{3}{8}$
• D. $\frac{3}{4}$

Answer 1

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

Let $E_1$ is the event of getting head in first toss and $E_2$ is the event of getting exactly two heads in three tosses then

$P\left(E_2/E_1\right)=\frac{P(E_2\cap E_1)}{P\left(E_1\right)}$

$P(E_2\cap E_1)=P(HHT,HTH)$

$=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{2}.\frac{1}{2}=\frac{1}{4}$

and $P\left(E_1\right)=\frac{1}{2}$

$\therefore P\left(E_2/E_1\right)=\frac{1/4}{1/2}=\frac{1}{2}$