A bag contains \(2 n+1\) coins. It is known that \(n\) of these coins have head on both sides whereas, the other \(n+1\) coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is \(\frac{31}{42}\), then the value of \(n\) is

Given, \(n\) coins have head on both the sides
and \((n+1)\) coins are fair coins.
Total coins \(=2 n+1\)
Let events \(E_1, E_2\) be the following
\(E_1=\) Event that an unfair coin is selected
\(E_2=\) Event that a fair coin is selected
\(\therefore \quad P\left(E_1\right)=\frac{n}{2 n+1}\) and \(P\left(E_2\right)=\frac{n+1}{2 n+1}\)
From the law of total probability,
\(\therefore \quad P(E)=P\left(E_1\right) \times P\left(\frac{E}{E_1}\right)+P\left(E_2\right) \times P\left(\frac{E}{E_2}\right)\)
\(\begin{array}{ll}\Rightarrow \quad & \frac{31}{42}=\frac{n}{2 n+1} \times 1+\frac{n+1}{2 n+1} \times \frac{1}{2} \\ \Rightarrow \quad & \frac{31}{42}=\frac{2 n+n+1}{2(2 n+1)} \\ \Rightarrow & 31 \times 2(2 n+1)=42(3 n+1) \\ \Rightarrow & 124 n+62=126 n+42 \\ \Rightarrow & 2 n=20 \\ & n=10\end{array}\)