Question

Three coins are tossed. If one of them shows tail, then the probability that all three coins show tail is

• A. $\frac{1}{7}$
• B. $\frac{1}{8}$
• C. $\frac{2}{7}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A

$\frac{1}{7}$

Explanation for the correct answer:

Given: Three coins are tossed

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

Let $\mathrm{A}$ be the event that all three coins show the tail.

Let $\mathrm{B}$ be the event that one coin show tail.

$\mathrm{A}=\left\{\mathrm{TTT}\right\}$

$\mathrm{B}=\left\{\text{TTT},\mathrm{HHT},\mathrm{HTH},\mathrm{THH},\mathrm{HTT},\mathrm{THT},\mathrm{TTH}\right\}=7$

$\mathrm{A}\cap \mathrm{B}=\left\{\mathrm{TTT}\right\}=1$

Probability $=\frac{Totalnumberoffavourableevents}{Totalpossibleevents}$

$P(A\cap B)=\frac{1}{8},P\left(B\right)=\frac{7}{8}$

Required probability $=\mathrm{P}(\mathrm{A}/\mathrm{B})$

Probability of event $\mathrm{A}$ when $\mathrm{B}$ is given.

$$\begin{gathered} \mathrm{P}(\mathrm{A}/\mathrm{B})=\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)} \\ =\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ =\frac{1}{7} \end{gathered}$$

Hence option A is the correct answer.