The probability that a particular day in the month of July is a rainy day is `3/4` . Two person whose credibility and `4/5` and `2/3` , respectively, claim that 15th July was a rainy day. Find the probability that it was really a rainy day.

Explanation

To solve the problem, we need to find the probability that it was really a rainy day given the claims made by two persons about July 15th. We will use Bayes' theorem for this calculation.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( R \) be the event that July 15th was a rainy day.
- Let \( A \) be the event that person A claims it was a rainy day.
- Let \( B \) be the event that person B claims it was a rainy day.

2. **Given Probabilities**:
- The probability that it is a rainy day: \( P(R) = \frac{3}{4} \)
- The probability that person A speaks the truth: \( P(A | R) = \frac{4}{5} \)
- The probability that person B speaks the truth: \( P(B | R) = \frac{2}{3} \)

The probabilities that they lie (speak false) are:
- \( P(A' | R) = 1 - P(A | R) = 1 - \frac{4}{5} = \frac{1}{5} \)
- \( P(B' | R) = 1 - P(B | R) = 1 - \frac{2}{3} = \frac{1}{3} \)

3. **Calculate the Joint Probabilities**:
- The probability that both A and B tell the truth when it is rainy:
\[ P(A \cap B | R) = P(A | R) \cdot P(B | R) = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15} \]

- The probability that both A and B lie when it is not rainy:
- The probability that it is not rainy: \( P(R') = 1 - P(R) = 1 - \frac{3}{4} = \frac{1}{4} \)
\[ P(A' \cap B' | R') = P(A' | R') \cdot P(B' | R') = \frac{1}{5} \cdot \frac{1}{3} = \frac{1}{15} \]

4. **Calculate the Total Probability of Claims**:
- The total probability that both A and B claim it was rainy:
\[ P(A \cap B) = P(A \cap B | R) \cdot P(R) + P(A' \cap B' | R') \cdot P(R') \]
Substituting the values:
\[ P(A \cap B) = \left(\frac{8}{15} \cdot \frac{3}{4}\right) + \left(\frac{1}{15} \cdot \frac{1}{4}\right) \]
\[ = \frac{8}{15} \cdot \frac{3}{4} + \frac{1}{15} \cdot \frac{1}{4} = \frac{6}{15} + \frac{1}{60} \]
Finding a common denominator (60):
\[ = \frac{24}{60} + \frac{1}{60} = \frac{25}{60} = \frac{5}{12} \]

5. **Final Calculation Using Bayes' Theorem**:
- We want \( P(R | A \cap B) \):
\[ P(R | A \cap B) = \frac{P(A \cap B | R) \cdot P(R)}{P(A \cap B)} \]
Substituting the values:
\[ P(R | A \cap B) = \frac{\frac{8}{15} \cdot \frac{3}{4}}{\frac{5}{12}} = \frac{\frac{6}{15}}{\frac{5}{12}} = \frac{6}{15} \cdot \frac{12}{5} = \frac{72}{75} = \frac{24}{25} \]

### Final Answer:
The probability that it was really a rainy day is \( \frac{24}{25} \).