A, B, C try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are `(3)/(4),(1)/(2),(5)/(8)` . The probability that target is hit by A or B but not C is

Options

A. `(21)/(64)`
B. `(7)/(8)`
C. `(7)/(32)`
D. `(9)/(64)`

Answer

To solve the problem, we need to find the probability that the target is hit by A or B but not by C. Let's break this down step by step.

### Step 1: Define the probabilities
Let:
- \( P(A_h) = \frac{3}{4} \) (Probability that A hits the target)
- \( P(B_h) = \frac{1}{2} \) (Probability that B hits the target)
- \( P(C_h) = \frac{5}{8} \) (Probability that C hits the target)

### Step 2: Calculate the probabilities of not hitting the target
Using the formula \( P(X_{nh}) = 1 - P(X_h) \):
- \( P(A_{nh}) = 1 - P(A_h) = 1 - \frac{3}{4} = \frac{1}{4} \)
- \( P(B_{nh}) = 1 - P(B_h) = 1 - \frac{1}{2} = \frac{1}{2} \)
- \( P(C_{nh}) = 1 - P(C_h) = 1 - \frac{5}{8} = \frac{3}{8} \)

### Step 3: Identify the required event
We need to find the probability that the target is hit by A or B but not by C. This can occur in three scenarios:
1. A hits, B does not hit, and C does not hit.
2. A does not hit, B hits, and C does not hit.
3. A hits, B hits, and C does not hit.

### Step 4: Calculate the probabilities for each scenario
1. **Scenario 1**: \( P(A_h) \cdot P(B_{nh}) \cdot P(C_{nh}) \)
\[ = P(A_h) \cdot P(B_{nh}) \cdot P(C_{nh}) = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{8} = \frac{3 \times 1 \times 3}{4 \times 2 \times 8} = \frac{9}{64} \]

2. **Scenario 2**: \( P(A_{nh}) \cdot P(B_h) \cdot P(C_{nh}) \)
\[ = P(A_{nh}) \cdot P(B_h) \cdot P(C_{nh}) = \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{3}{8} = \frac{1 \times 1 \times 3}{4 \times 2 \times 8} = \frac{3}{64} \]

3. **Scenario 3**: \( P(A_h) \cdot P(B_h) \cdot P(C_{nh}) \)
\[ = P(A_h) \cdot P(B_h) \cdot P(C_{nh}) = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{3}{8} = \frac{3 \times 1 \times 3}{4 \times 2 \times 8} = \frac{9}{64} \]

### Step 5: Add the probabilities of all scenarios
Now, we sum the probabilities from all three scenarios:
\[ P(A \text{ or } B \text{ but not } C) = \frac{9}{64} + \frac{3}{64} + \frac{9}{64} = \frac{21}{64} \]

### Final Answer
The probability that the target is hit by A or B but not by C is \( \frac{21}{64} \).

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