Three persons A, B and C shoot to hit a target. If in trials, A hits the target 4 times in 5 shots, B hits 3 times in 4 shots and C hits 2 times in 3 trials. Find the probability that: Exactly two persons hit the target.

Answer

To find the probability that exactly two persons hit the target among A, B, and C, we will follow these steps:

### Step 1: Determine the probabilities of hitting and missing the target for each person.

- **Person A** hits the target 4 times in 5 shots:
\[ P(A \text{ hits}) = \frac{4}{5} \]
\[ P(A \text{ misses}) = 1 - P(A \text{ hits}) = 1 - \frac{4}{5} = \frac{1}{5} \]

- **Person B** hits the target 3 times in 4 shots:
\[ P(B \text{ hits}) = \frac{3}{4} \]
\[ P(B \text{ misses}) = 1 - P(B \text{ hits}) = 1 - \frac{3}{4} = \frac{1}{4} \]

- **Person C** hits the target 2 times in 3 shots:
\[ P(C \text{ hits}) = \frac{2}{3} \]
\[ P(C \text{ misses}) = 1 - P(C \text{ hits}) = 1 - \frac{2}{3} = \frac{1}{3} \]

### Step 2: Identify the scenarios where exactly two persons hit the target.

The scenarios where exactly two persons hit the target are:
1. A and B hit, C misses.
2. A and C hit, B misses.
3. B and C hit, A misses.

### Step 3: Calculate the probabilities for each scenario.

1. **Probability that A and B hit, C misses**:
\[ P(A \text{ hits}) \times P(B \text{ hits}) \times P(C \text{ misses}) = \frac{4}{5} \times \frac{3}{4} \times \frac{1}{3} \]
\[ = \frac{4 \times 3 \times 1}{5 \times 4 \times 3} = \frac{1}{5} \]

2. **Probability that A and C hit, B misses**:
\[ P(A \text{ hits}) \times P(B \text{ misses}) \times P(C \text{ hits}) = \frac{4}{5} \times \frac{1}{4} \times \frac{2}{3} \]
\[ = \frac{4 \times 1 \times 2}{5 \times 4 \times 3} = \frac{8}{60} = \frac{2}{15} \]

3. **Probability that B and C hit, A misses**:
\[ P(A \text{ misses}) \times P(B \text{ hits}) \times P(C \text{ hits}) = \frac{1}{5} \times \frac{3}{4} \times \frac{2}{3} \]
\[ = \frac{1 \times 3 \times 2}{5 \times 4 \times 3} = \frac{6}{60} = \frac{1}{10} \]

### Step 4: Sum the probabilities of all scenarios.

Now, we sum the probabilities of all three scenarios:
\[ P(\text{exactly 2 hit}) = P(A \text{ and } B \text{ hit, } C \text{ misses}) + P(A \text{ and } C \text{ hit, } B \text{ misses}) + P(B \text{ and } C \text{ hit, } A \text{ misses}) \]
\[ = \frac{1}{5} + \frac{2}{15} + \frac{1}{10} \]

### Step 5: Find a common denominator and add.

The least common multiple of 5, 15, and 10 is 30. We convert each fraction:
\[ \frac{1}{5} = \frac{6}{30}, \quad \frac{2}{15} = \frac{4}{30}, \quad \frac{1}{10} = \frac{3}{30} \]
Now, we add them:
\[ P(\text{exactly 2 hit}) = \frac{6}{30} + \frac{4}{30} + \frac{3}{30} = \frac{13}{30} \]

### Final Answer:
The probability that exactly two persons hit the target is:
\[ \boxed{\frac{13}{30}} \]