Suppose a girl throws a die. If she gets an even number, she tosses a coin three times and notes the number of 'tails'. If she gets an odd number, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', what is the probability that she threw odd numbers with the die ?

Answer

To solve the problem, we need to find the probability that the girl threw an odd number on the die given that she obtained exactly one tail. We will use Bayes' theorem to find this probability.

### Step 1: Define Events
Let:
- \( A \): The event that the girl throws an odd number on the die.
- \( B \): The event that she obtains exactly one tail.

We want to find \( P(A|B) \), the probability of event \( A \) given event \( B \).

### Step 2: Use Bayes' Theorem
According to Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]

### Step 3: Calculate \( P(A) \)
The probability of throwing an odd number on a die:
\[ P(A) = \frac{3}{6} = \frac{1}{2} \]
(since the odd numbers on a die are 1, 3, and 5).

### Step 4: Calculate \( P(B|A) \)
If she throws an odd number, she tosses a coin once. The outcomes are either heads (H) or tails (T). The probability of getting exactly one tail (which is the only outcome possible in this case) is:
\[ P(B|A) = \frac{1}{2} \]

### Step 5: Calculate \( P(B) \)
To find \( P(B) \), we need to consider both cases: when she throws an even number and when she throws an odd number.

1. **Case 1: Even Number**
- Probability of throwing an even number:
\[ P(\text{Even}) = \frac{3}{6} = \frac{1}{2} \]
- If she throws an even number, she tosses a coin three times. The probability of getting exactly one tail in three tosses is calculated using the binomial distribution:
\[ P(B|\text{Even}) = \binom{3}{1} \left(\frac{1}{2}\right)^1 \left(\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{3}{8} \]

2. **Case 2: Odd Number**
- We already calculated:
\[ P(B|A) = \frac{1}{2} \]

Now, we can calculate \( P(B) \):
\[ P(B) = P(B|\text{Even}) \cdot P(\text{Even}) + P(B|A) \cdot P(A) \]
\[ P(B) = \left(\frac{3}{8} \cdot \frac{1}{2}\right) + \left(\frac{1}{2} \cdot \frac{1}{2}\right) \]
\[ P(B) = \frac{3}{16} + \frac{1}{4} = \frac{3}{16} + \frac{4}{16} = \frac{7}{16} \]

### Step 6: Substitute Values into Bayes' Theorem
Now we can substitute all the values back into Bayes' theorem:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} = \frac{\left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)}{\frac{7}{16}} \]
\[ P(A|B) = \frac{\frac{1}{4}}{\frac{7}{16}} = \frac{1}{4} \cdot \frac{16}{7} = \frac{4}{7} \]

### Final Answer
Thus, the probability that she threw an odd number given that she obtained exactly one tail is:
\[ \boxed{\frac{4}{7}} \]