Question

A bag contains \(n+1\) coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is \(\frac{7}{12}\), then the value of \(n\) is

• A. \(3\)
• B. \(4\)
• C. \(5\)
• D. None of these

Hint:

Use total probability when selection precedes an experiment.

Answer 1

The Correct Option is C

Step 1: Probability of selecting double-headed coin \(=\frac{1}{n+1}\).
Step 2: \[ P(H)=\frac{1}{n+1}\cdot1+\frac{n}{n+1}\cdot\frac{1}{2} =\frac{n+2}{2(n+1)} \]
Step 3: \[ \frac{n+2}{2(n+1)}=\frac{7}{12}\Rightarrow n=5 \]