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 (31)/(42), then n is equal to

• A. \(10\)
• B. \(11\)
• C. \(12\)
• D. 13

Hint:

Use total probability when multiple types of coins are involved.

Answer 1

The Correct Option is C

Probability of choosing a double-headed coin \( = \dfrac{n}{2n+1} \).
Probability of choosing a fair coin \( = \dfrac{n+1}{2n+1} \).

\( P(H) = \dfrac{n}{2n+1} \cdot 1 + \dfrac{n+1}{2n+1} \cdot \dfrac{1}{2} \)
\( = \dfrac{3n+1}{2(2n+1)} \)
Given:
\( \dfrac{3n+1}{2(2n+1)} = \dfrac{31}{42} \)
\( 42(3n+1) = 31 \cdot 2(2n+1) \Rightarrow 126n + 42 = 62(2n+1) \)
\( 126n + 42 = 124n + 62 \)
\( 2n = 20 \Rightarrow n = 10 \)