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

Let $E$ be the event that the selected coin shows heads. Let $C_1$ be the event that the two-headed coin is selected, and $C_2$ be the event that a fair coin is selected.

The probability of selecting the two-headed coin is $P(C_1) = \dfrac{1}{n+1}$.

The probability of selecting a fair coin is $P(C_2) = \dfrac{n}{n+1}$.

The probability of getting heads given the two-headed coin is $P(E|C_1) = 1$.

The probability of getting heads given a fair coin is $P(E|C_2) = \dfrac{1}{2}$.

Using the law of total probability, $P(E) = P(E|C_1)P(C_1) + P(E|C_2)P(C_2)$.

Substituting the known values, $\dfrac{7}{12} = (1)\left(\dfrac{1}{n+1}\right) + \left(\dfrac{1}{2}\right)\left(\dfrac{n}{n+1}\right)$.

Multiplying by $2(n+1)$, we get $\dfrac{7}{12} \times 2(n+1) = 2 + n$.

$\dfrac{7(n+1)}{6} = n + 2$.

$7n + 7 = 6n + 12$.

$n = 5$.

Answer: 5