Question

Three fair coins are tossed. If both heads and tails appear, then the probability that exactly one head appears is

• A. $\frac{3}{8}$
• B. $\frac{1}{6}$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$

Answer 1

The correct option is C

$\frac{1}{2}$

Explanation for the correct answer:

Given: Three coins are tossed

Sample space $=\left\{\mathrm{HHH},\mathrm{HTH},\mathrm{HHT},\mathrm{THH},\mathrm{TTH},\mathrm{THT},\mathrm{HTT},\mathrm{TTT}\right\},\mathrm{n}\left(\mathrm{S}\right)=8$

The number of outcomes where both head and tail appear $=\left\{\mathrm{HTH},\mathrm{HHT},\mathrm{THH},\mathrm{TTH},\mathrm{HTT},\mathrm{THT}\right\},\mathrm{n}\left(\mathrm{A}\right)=6$

The number of outcomes where exactly one head appears $=\left\{\mathrm{HTH},\mathrm{THH},\mathrm{HHT}\right\},\mathrm{n}\left(\mathrm{B}\right)=3$

We have to find the probability of $B$when the probability of $A$is given.

$(A\cap B)=3,P(A\cap B)=\frac{3}{8}$

$$\begin{gathered} \mathrm{P}\left(\mathrm{B}\right|\mathrm{A})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)} \\ \mathrm{P}(\mathrm{B}\left|\mathrm{A}\right)=\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{{\displaystyle \raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ \mathrm{P}\left(\mathrm{B}\right|\mathrm{A})=\frac{3}{6}=\frac{1}{2} \end{gathered}$$

Hence option C is the correct answer.