The probability that a police inspector Ravi will catch a thief in a day is `1/4` and the probability he will catch a robber in that day is `1/5` and the probability that he will catch both a thief and a robber in a day is `1/15` then what is the probability that Ravi will catch at least 1 mischief?

Options

A. `23//60`
B. `19//60`
C. `7//20`
D. None of these

Answer

To solve the problem, we need to find the probability that Ravi will catch at least one mischief (either a thief or a robber or both). We will use the formula for the union of two events in probability.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( A \) be the event that Ravi catches a thief.
- Let \( B \) be the event that Ravi catches a robber.

2. **Given Probabilities**:
- The probability that Ravi catches a thief:
\[ P(A) = \frac{1}{4} \]
- The probability that Ravi catches a robber:
\[ P(B) = \frac{1}{5} \]
- The probability that Ravi catches both a thief and a robber:
\[ P(A \cap B) = \frac{1}{15} \]

3. **Use the Formula for Union of Two Events**:
The probability that Ravi catches at least one mischief (either a thief or a robber or both) is given by:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

4. **Substitute the Values**:
Now, substitute the values into the formula:
\[ P(A \cup B) = \frac{1}{4} + \frac{1}{5} - \frac{1}{15} \]

5. **Find a Common Denominator**:
The least common multiple (LCM) of 4, 5, and 15 is 60. Convert each fraction:
- \( P(A) = \frac{1}{4} = \frac{15}{60} \)
- \( P(B) = \frac{1}{5} = \frac{12}{60} \)
- \( P(A \cap B) = \frac{1}{15} = \frac{4}{60} \)

6. **Calculate the Union**:
Now substitute these values back into the equation:
\[ P(A \cup B) = \frac{15}{60} + \frac{12}{60} - \frac{4}{60} \]
Combine the fractions:
\[ P(A \cup B) = \frac{15 + 12 - 4}{60} = \frac{23}{60} \]

7. **Final Result**:
Therefore, the probability that Ravi will catch at least one mischief is:
\[ \boxed{\frac{23}{60}} \]