A bag contains $(n + 1)$ coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $7/12,$ then the value of $n$ is

Let $E_1$ denote the event "a coin with two heads is selected and $E_2$ denote the event "a fair coin is selected". Let A be the event "the toss results in head". Then,

                      $$\begin{gathered} P\left(E_1\right)=\frac{1}{n+1},P\left(E_2\right)=\frac{n}{n+1}, \\ P\left(A/E_1\right)=1 \end{gathered}$$

and    $P\left(A/E_2\right)=1/2.$

$\therefore P\left(A\right)=P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)$

$\Rightarrow \frac{7}{12}=\frac{1}{n+1}\times 1+\frac{n}{n+1}\times \frac{1}{2}[\because P(A)=7/12]$

$\Rightarrow 12+6n=7n+7\Rightarrow n=5$