A box contains N coins, m of which are fair and the rest are biased. The probability of getting a head when a fair coin is tossed, is 1/2 while it is 2/3 when a biased coin is tossed. A coin is drawn from the box at random and is tossed twice. The first time it shows head and the second time it shows tail. What is the probability that the coin drawn is fair ?

Answer

To solve the problem, we will use Bayes' theorem. Let's denote the events as follows:

- \( E_1 \): The event that the selected coin is fair.
- \( E_2 \): The event that the selected coin is biased.
- \( A \): The event that the first toss shows heads and the second toss shows tails.

We need to find \( P(E_1 | A) \), the probability that the coin drawn is fair given that the first toss is heads and the second toss is tails.

### Step 1: Calculate \( P(E_1) \) and \( P(E_2) \)

- The probability of selecting a fair coin:
\[ P(E_1) = \frac{m}{N} \]

- The probability of selecting a biased coin:
\[ P(E_2) = \frac{N - m}{N} \]

### Step 2: Calculate \( P(A | E_1) \) and \( P(A | E_2) \)

- If the selected coin is fair, the probability of getting heads on the first toss and tails on the second toss is:
\[ P(A | E_1) = P(\text{Head on 1st toss} | E_1) \times P(\text{Tail on 2nd toss} | E_1) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

- If the selected coin is biased, the probability of getting heads on the first toss and tails on the second toss is:
\[ P(A | E_2) = P(\text{Head on 1st toss} | E_2) \times P(\text{Tail on 2nd toss} | E_2) = \frac{2}{3} \times \left(1 - \frac{2}{3}\right) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} \]

### Step 3: Apply Bayes' Theorem

According to Bayes' theorem:
\[ P(E_1 | A) = \frac{P(A | E_1) \cdot P(E_1)}{P(A)} \]

Where \( P(A) \) can be calculated using the law of total probability:
\[ P(A) = P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2) \]

Now substituting the values we calculated:
\[ P(A) = \left(\frac{1}{4} \cdot \frac{m}{N}\right) + \left(\frac{2}{9} \cdot \frac{N - m}{N}\right) \]

### Step 4: Substitute into Bayes' Theorem

Now substituting \( P(A) \) back into Bayes' theorem:
\[ P(E_1 | A) = \frac{\left(\frac{1}{4} \cdot \frac{m}{N}\right)}{\left(\frac{1}{4} \cdot \frac{m}{N}\right) + \left(\frac{2}{9} \cdot \frac{N - m}{N}\right)} \]

### Step 5: Simplify the Expression

To simplify, we can multiply both the numerator and denominator by \( 36N \) (the least common multiple of denominators):
\[ P(E_1 | A) = \frac{9m}{9m + 8(N - m)} = \frac{9m}{8N + m} \]

### Final Answer

Thus, the probability that the coin drawn is fair given that the first toss is heads and the second toss is tails is:
\[ P(E_1 | A) = \frac{9m}{8N + m} \]