Question

If there are (n + 1) coins of which n coins are fair and one is double headed. A coin is randomly selected and tossed, if probability that head occurs is \( \frac{9}{16} \) then n is:

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem uses the Law of Total Probability. The event "Head occurs" (\( H \)) can happen in two ways: choosing a fair coin (\( F \)) or choosing the double-headed coin (\( D \)).
Step 2: Key Formula or Approach:
\[ P(H) = P(F)P(H|F) + P(D)P(H|D) \]
Step 3: Detailed Explanation:
1. Total coins = \( n + 1 \). 2. Probability of picking a fair coin \( P(F) = \frac{n}{n+1} \). 3. Probability of picking the double-headed coin \( P(D) = \frac{1}{n+1} \). 4. \( P(H|F) = 1/2 \) (Fair coin) and \( P(H|D) = 1 \) (Double-headed). 5. Setting up the equation: \[ \frac{n}{n+1} \cdot \frac{1}{2} + \frac{1}{n+1} \cdot 1 = \frac{9}{16} \] \[ \frac{n/2 + 1}{n+1} = \frac{9}{16} \implies \frac{n+2}{2(n+1)} = \frac{9}{16} \] 6. Cross-multiply: \[ 16(n + 2) = 18(n + 1) \] \[ 16n + 32 = 18n + 18 \implies 2n = 14 \implies n = 7 \]
Step 4: Final Answer:
The value of \( n \) is 7.