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:

$$P(H) = P(F) \cdot P(H|F) + P(B) \cdot P(H|B)$$

Probability of choosing a fair coin: $$P(F) = \frac{N}{N+1}$$

Probability of choosing the two-headed coin: $$P(B) = \frac{1}{N+1}$$

$$P(H|F) = \frac{1}{2}$$

$$P(H|B) = 1$$

Given that $$P(H) = \frac{9}{16}$$:

$$\frac{9}{16} = \left(\frac{N}{N+1} \times \frac{1}{2}\right) + \left(\frac{1}{N+1} \times 1\right)$$

$$\frac{9}{16} = \frac{N}{2(N+1)} + \frac{1}{N+1}$$

$$\frac{9}{16} = \frac{N + 2}{2(N+1)}$$

$$2N = 14 \implies N = 7$$