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 \[10\].