Question

A bag contains $(N + 1)$ coins - $N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then $N$ is equal to:

• A. 5
• B. 7
• C. 8
• D. 9

Hint:

Use the law of total probability: $P(H) = P(H|Fair)P(Fair) + P(H|Biased)P(Biased)$.

Answer 1

The Correct Option is B

We can visualize the problem in terms of the total outcomes. In a bag of $N+1$ coins, there are $N$ coins with outcomes {H, T} and 1 coin with outcomes {H, H}.
The total number of equally likely outcomes when a coin is picked and tossed is the sum of the sides of all coins, but we are picking coins with different head-probabilities. The probability of choosing a fair coin is $\frac{N}{N+1}$ and the two-headed coin is $\frac{1}{N+1}$.
Expected Head Probability = $\text{Prob(fair)} \times \text{Prob(H|fair)} + \text{Prob(biased)} \times \text{Prob(H|biased)}$.
$$ \frac{9}{16} = \frac{N}{N+1} \cdot \frac{1}{2} + \frac{1}{N+1} \cdot 1 $$
Multiplying both sides by $16 \times 2(N+1)$ to clear denominators:
$$ 9 \cdot 2(N+1) = 16N + 32 $$
$$ 18N + 18 = 16N + 32 $$
$$ 2N = 14 \implies N = 7 $$
Thus, there are 7 fair coins in the bag.