Determine Number of Fair Coins in Bag Probability Problem

To find the value of $N$, we can use the Law of Total Probability.

Let the events be defined as follows:

• $E_1$: The event that a fair coin is selected from the bag.
• $E_2$: The event that the two-headed coin is selected from the bag.
• $H$: The event of getting a 'Head' on tossing the chosen coin.

The bag contains a total of $(N + 1)$ coins, out of which $N$ are fair coins and $1$ is a two-headed coin.

The probabilities of selecting these coins are: $P(E_1) = \frac{N}{N + 1}$ $P(E_2) = \frac{1}{N + 1}$

The conditional probabilities of getting a 'Head' given the type of coin chosen are:

• For a fair coin: $P(H \mid E_1) = \frac{1}{2}$
• For the two-headed coin: $P(H \mid E_2) = 1$

Using the Law of Total Probability, the overall probability of getting a 'Head' is given by: $P(H) = P(E_1) \cdot P(H \mid E_1) + P(E_2) \cdot P(H \mid E_2)$

Substitute the respective values into the formula: $P(H) = \left(\frac{N}{N + 1}\right) \cdot \frac{1}{2} + \left(\frac{1}{N + 1}\right) \cdot 1$ $P(H) = \frac{N + 2}{2(N + 1)}$

It is given that the probability of getting 'Head' is $\frac{9}{16}$. Equating this to our expression: $\frac{N + 2}{2(N + 1)} = \frac{9}{16}$

Cross-multiplying gives: $\frac{N + 2}{N + 1} = \frac{9}{8}$ $8(N + 2) = 9(N + 1)$ $8N + 16 = 9N + 9$ $9N - 8N = 16 - 9$ $N = 7$

Thus, the value of $N$ is equal to 7, which corresponds to Option B.