Three persons A , B and C shoot to hit a target. Their probabilities of hitting the target are `5/(6), 4/(5) and 3/(4)` respectively. Find the probability that :
Exactly two persons hit the target.

Answer

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

1. **Identify the probabilities of hitting and missing the target:**
- Probability that A hits the target, \( P(A) = \frac{5}{6} \)
- Probability that B hits the target, \( P(B) = \frac{4}{5} \)
- Probability that C hits the target, \( P(C) = \frac{3}{4} \)

The probabilities of missing the target are:
- Probability that A misses the target, \( P(A') = 1 - P(A) = 1 - \frac{5}{6} = \frac{1}{6} \)
- Probability that B misses the target, \( P(B') = 1 - P(B) = 1 - \frac{4}{5} = \frac{1}{5} \)
- Probability that C misses the target, \( P(C') = 1 - P(C) = 1 - \frac{3}{4} = \frac{1}{4} \)

2. **Calculate the probabilities for the scenarios where exactly two persons hit the target:**
- Scenario 1: A and B hit the target, C misses.
\[ P(A \cap B \cap C') = P(A) \cdot P(B) \cdot P(C') = \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} \]
- Scenario 2: A and C hit the target, B misses.
\[ P(A \cap B' \cap C) = P(A) \cdot P(B') \cdot P(C) = \frac{5}{6} \cdot \frac{1}{5} \cdot \frac{3}{4} \]
- Scenario 3: B and C hit the target, A misses.
\[ P(A' \cap B \cap C) = P(A') \cdot P(B) \cdot P(C) = \frac{1}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} \]

3. **Calculate each scenario:**
- For Scenario 1:
\[ P(A \cap B \cap C') = \frac{5}{6} \cdot \frac{4}{5} \cdot \frac{1}{4} = \frac{5 \cdot 4 \cdot 1}{6 \cdot 5 \cdot 4} = \frac{1}{6} \]
- For Scenario 2:
\[ P(A \cap B' \cap C) = \frac{5}{6} \cdot \frac{1}{5} \cdot \frac{3}{4} = \frac{5 \cdot 1 \cdot 3}{6 \cdot 5 \cdot 4} = \frac{3}{24} = \frac{1}{8} \]
- For Scenario 3:
\[ P(A' \cap B \cap C) = \frac{1}{6} \cdot \frac{4}{5} \cdot \frac{3}{4} = \frac{1 \cdot 4 \cdot 3}{6 \cdot 5 \cdot 4} = \frac{3}{30} = \frac{1}{10} \]

4. **Sum the probabilities of all scenarios:**
\[ P(\text{Exactly 2 hit}) = P(A \cap B \cap C') + P(A \cap B' \cap C) + P(A' \cap B \cap C) \]
\[ = \frac{1}{6} + \frac{1}{8} + \frac{1}{10} \]

5. **Find a common denominator and sum:**
The least common multiple of 6, 8, and 10 is 120.
- Convert each fraction:
\[ \frac{1}{6} = \frac{20}{120}, \quad \frac{1}{8} = \frac{15}{120}, \quad \frac{1}{10} = \frac{12}{120} \]
- Now sum them:
\[ P(\text{Exactly 2 hit}) = \frac{20 + 15 + 12}{120} = \frac{47}{120} \]

Thus, the probability that exactly two persons hit the target is \( \frac{47}{120} \).