Two men hit at a target with probabilities `(1)/(2)` and `(1)/(3)` respectively. What is the probability that exactly one of them hits the target?

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(6)`
D. `(2)/(3)`

Answer

To find the probability that exactly one of the two men hits the target, we can follow these steps:

### Step 1: Define the probabilities
Let:
- The probability that man A hits the target, \( P(A) = \frac{1}{2} \)
- The probability that man B hits the target, \( P(B) = \frac{1}{3} \)

### Step 2: Calculate the probabilities of missing the target
The probability that man A misses the target is:
\[ P(A') = 1 - P(A) = 1 - \frac{1}{2} = \frac{1}{2} \]

The probability that man B misses the target is:
\[ P(B') = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 3: Determine the scenarios for exactly one hit
There are two scenarios where exactly one of them hits the target:
1. Man A hits and man B misses.
2. Man A misses and man B hits.

### Step 4: Calculate the probability for each scenario
**Scenario 1: A hits and B misses**
The probability for this scenario is:
\[ P(A \text{ hits and } B \text{ misses}) = P(A) \cdot P(B') = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3} \]

**Scenario 2: A misses and B hits**
The probability for this scenario is:
\[ P(A' \text{ misses and } B \text{ hits}) = P(A') \cdot P(B) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \]

### Step 5: Add the probabilities of both scenarios
To find the total probability that exactly one of them hits the target, we add the probabilities of the two scenarios:
\[ P(\text{exactly one hits}) = P(A \text{ hits and } B \text{ misses}) + P(A' \text{ misses and } B \text{ hits}) \]
\[ P(\text{exactly one hits}) = \frac{1}{3} + \frac{1}{6} \]

### Step 6: Find a common denominator and simplify
To add \( \frac{1}{3} \) and \( \frac{1}{6} \), we can convert \( \frac{1}{3} \) to have a denominator of 6:
\[ \frac{1}{3} = \frac{2}{6} \]
So,
\[ P(\text{exactly one hits}) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

### Final Answer
The probability that exactly one of them hits the target is:
\[ \frac{1}{2} \]

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