Probabilities that A, B, C can hit a target are `1/3, 1/4, 1/5` respectively. If they all fire simultaneously, the probability that exactly two of them hit the target is

Options

A. `1/20`
B. `1/10`
C. `3/20`
D. `3/10`

Answer

To find the probability that exactly two out of A, B, and C hit the target, we can break down the problem into manageable steps.

### Step-by-Step Solution:

1. **Identify the probabilities**:
- Probability that A hits the target, \( P(A) = \frac{1}{3} \)
- Probability that B hits the target, \( P(B) = \frac{1}{4} \)
- Probability that C hits the target, \( P(C) = \frac{1}{5} \)

2. **Calculate the probabilities of missing the target**:
- Probability that A misses the target, \( P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \)
- Probability that B misses the target, \( P(B') = 1 - P(B) = 1 - \frac{1}{4} = \frac{3}{4} \)
- Probability that C misses the target, \( P(C') = 1 - P(C) = 1 - \frac{1}{5} = \frac{4}{5} \)

3. **Determine the scenarios for exactly two hits**:
The scenarios where exactly two of them hit the target are:
- A and B hit, C misses: \( P(A) \cdot P(B) \cdot P(C') \)
- A and C hit, B misses: \( P(A) \cdot P(B') \cdot P(C) \)
- B and C hit, A misses: \( P(A') \cdot P(B) \cdot P(C) \)

4. **Calculate the probabilities for each scenario**:
- For A and B hit, C misses:
\[ P(A) \cdot P(B) \cdot P(C') = \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{4}{5} = \frac{1 \cdot 1 \cdot 4}{3 \cdot 4 \cdot 5} = \frac{4}{60} = \frac{1}{15} \]
- For A and C hit, B misses:
\[ P(A) \cdot P(B') \cdot P(C) = \frac{1}{3} \cdot \frac{3}{4} \cdot \frac{1}{5} = \frac{1 \cdot 3 \cdot 1}{3 \cdot 4 \cdot 5} = \frac{3}{60} = \frac{1}{20} \]
- For B and C hit, A misses:
\[ P(A') \cdot P(B) \cdot P(C) = \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} = \frac{2 \cdot 1 \cdot 1}{3 \cdot 4 \cdot 5} = \frac{2}{60} = \frac{1}{30} \]

5. **Sum the probabilities of all scenarios**:
\[ P(\text{exactly 2 hits}) = P(A \text{ and } B \text{ hit, } C \text{ misses}) + P(A \text{ and } C \text{ hit, } B \text{ misses}) + P(B \text{ and } C \text{ hit, } A \text{ misses} \]
\[ = \frac{1}{15} + \frac{1}{20} + \frac{1}{30} \]

6. **Find a common denominator**:
The least common multiple of 15, 20, and 30 is 60.
- Convert each fraction:
\[ \frac{1}{15} = \frac{4}{60}, \quad \frac{1}{20} = \frac{3}{60}, \quad \frac{1}{30} = \frac{2}{60} \]

7. **Add the fractions**:
\[ P(\text{exactly 2 hits}) = \frac{4}{60} + \frac{3}{60} + \frac{2}{60} = \frac{9}{60} = \frac{3}{20} \]

### Final Answer:
The probability that exactly two of A, B, and C hit the target is \( \frac{3}{20} \).