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

Given, $n$ coins have head on both sides and $(n + 1)$ coins are fair coins.

Let ${E_1} =$ Event that an unfair coin is selected

${E_2} =$ Event that a fair coin is selected

$E =$ Event that the toss results in a head

$\therefore$ $P\left( {{E_1}} \right) = \frac{n}{{2n + 1}}$

and $P\left( {{E_2}} \right) = \frac{{n + 1}}{{2n + 1}}$

Also, $P\left( {\frac{E}{{{E_1}}}} \right) = 1$

and $P\left( {\frac{E}{{{E_2}}}} \right) = \frac{1}{2}$

$\therefore$ $P(E) = P\left( {{E_1}} \right) \cdot P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right) \cdot P\left( {\frac{E}{{{E_2}}}} \right) = \frac{n}{{2n + 1}} \cdot 1 + \frac{{n + 1}}{{2n + 1}} \cdot \frac{1}{2}$

$\Rightarrow$ $\frac{{31}}{{42}} = \frac{{2n + n + 1}}{{2(2n + 1)}} \Rightarrow \frac{{31}}{{42}} = \frac{{3n + 1}}{{4n + 2}}$

$\Rightarrow$ $124n + 62 = 126n + 42$
$\Rightarrow$ $2n = 20 \Rightarrow n = 10$