Question:

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

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Use total probability = sum of (probability × conditional probability).
Updated On: Mar 23, 2026
  • \(10\)
  • \(11\)
  • \(12\)
  • \(13\)
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The Correct Option is B

Solution and Explanation


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: \[ \frac{3n+1}{2(2n+1)}=\frac{31}{42} \Rightarrow n=11 \]
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