Mathematics - Probability Question with Solution | TestHub

Total number of coins $=2\mathrm{n}+1$ Consider the following events: ${\mathrm{E}}_1=$ Getting a coin having head on both sides from the bag. ${\mathrm{E}}_2=$ Getting a fair coin from the bag $\mathrm{A}=$ Toss results in a head Given: $\mathrm{P}(\mathrm{A})=\frac{31}{42},\mathrm{P}\left({\mathrm{E}}_1\right)=\frac{\mathrm{n}}{2\mathrm{n}+1}$ and $\mathrm{P}\left({\mathrm{E}}_2\right)=\frac{\mathrm{n}+1}{2\mathrm{n}+1}$ Then, $\mathrm{P}(\mathrm{A})=\mathrm{P}\left({\mathrm{E}}_1\right)\mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_1\right)+\mathrm{P}\left({\mathrm{E}}_2\right)\mathrm{P}\left(\mathrm{A}/{\mathrm{E}}_2\right)$ $\Rightarrow \frac{31}{42}=\frac{n}{2n+1}\times 1+\frac{n+1}{2n+1}\times \frac{1}{2}$ $\frac{31}{42}=\frac{n}{2n+1}+\frac{n+1}{2(2n+1)}$ $\Rightarrow \frac{31}{42}=\frac{3n+1}{2(2n+1)}$ $\Rightarrow \frac{31}{21}=\frac{3n+1}{2n+1}$ $n=10$