A bag contains \((2n+1)\) coins. It is known that \(n\) of these coins have a head on both sides, whereas the remaining \((n+1)\) coins are fair. A coin is picked up at random from the bag and tossed. If the probability that the toss results in a head is \(\frac{31}{42}\), then \(n\) is equal to
Show Hint
Use total probability = sum of (probability × conditional probability).
Step 1: Probability of choosing a double-headed coin:
\[
\frac{n}{2n+1}
\]
Step 2: Probability of choosing a fair coin:
\[
\frac{n+1}{2n+1}
\]
Step 3: Total probability of head:
\[
P(H)=\frac{n}{2n+1}\cdot1+\frac{n+1}{2n+1}\cdot\frac12
=\frac{3n+1}{2(2n+1)}
\]
Step 4: Given:
\[
\frac{3n+1}{2(2n+1)}=\frac{31}{42}
\Rightarrow n=11
\]