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

This problem involves the law of total probability. Let $E_1$ be the event of selecting a fair coin and $E_2$ be the event of selecting the two-headed coin. Let $A$ be the event of getting a 'Head'.
The total number of coins is $N + 1$.
The probability of choosing a fair coin is $P(E_1) = \frac{N}{N+1}$.
The probability of choosing the two-headed coin is $P(E_2) = \frac{1}{N+1}$.
If a fair coin is selected, the probability of getting a head is $P(A|E_1) = \frac{1}{2}$.
If the two-headed coin is selected, the probability of getting a head is $P(A|E_2) = 1$.
According to the law of total probability:
$$ P(A) = P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2) $$
Substituting the given values:
$$ \frac{9}{16} = \left( \frac{N}{N+1} \right) \left( \frac{1}{2} \right) + \left( \frac{1}{N+1} \right) (1) $$
$$ \frac{9}{16} = \frac{N}{2(N+1)} + \frac{1}{N+1} $$
Taking the common denominator on the right side:
$$ \frac{9}{16} = \frac{N + 2}{2(N+1)} $$
Cross-multiplying:
$$ 18(N + 1) = 16(N + 2) $$
$$ 18N + 18 = 16N + 32 $$
$$ 2N = 14 \implies N = 7 $$