A bag contains \[(2n+1)\] coins. It is known that n of these coins has 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 \[31/42\], determine the value of n.

Given, n coins are two headed coins and the remaining \[(n+1)\] coins are fair.

Let  \[{{E}_{1}}\] : the event that unfair coin is selected

\[{{E}_{2}}\] : the event that the fair coin is selected

E: the event that the toss results in a head

So,

\[P({{E}_{1}})\text{ }=\text{ }n/\left( 2n\text{ }+\text{ }1 \right)\text{ }and\text{ }P({{E}_{2}})\text{ }=\text{ }\left( n\text{ }+\text{ }1 \right)/\text{ }\left( 2n\text{ }+1 \right)\]

\[P(E/{{E}_{1}})\text{ }=\text{ }1\] (As it’s a sure event)

\[P(E/{{E}_{2}})\text{ }=\text{ 1/2}\]

Therefore, the required value of n is \[10\].