Question

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 the toss results in heads is $\frac{7}{12}$, then the value of $n$ is

Answer 1

Let $E_1$ denote the event of selecting the coin with two heads,
and $E_2$ denote the event of selecting a fair coin.
Then
$P\left(E_1\right)=\frac{1}{n+1},P\left(E_2\right)=\frac{n}{n+1}$
Let $A$ denote the event that the toss results in heads.
Then, by total probability theorem,
$P\left(A\right)=P\left(E_1\right)P\left(A|E_1\right)+P\left(E_2\right)P\left(A|E_2\right)$
$\Rightarrow \frac{7}{12}=\frac{1}{n+1}\times 1+\frac{n}{n+1}\times \frac{1}{2}\Rightarrow \frac{7}{6}=\frac{n+2}{n+1}$
$\Rightarrow n=5$