Two independent events A and B have `P(A)= (1)/(3)` and `P(B)= (3)/(4)` What is the probability that exactly one of the two events A or B occurs?

Options

A. `(1)/(4)`
B. `(5)/(6)`
C. `(5)/(12)`
D. `(7)/(12)`

Answer

To find the probability that exactly one of the two independent events A or B occurs, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the probabilities of events A and B:**
- Given: \( P(A) = \frac{1}{3} \)
- Given: \( P(B) = \frac{3}{4} \)

2. **Calculate the probabilities of the complements of events A and B:**
- The probability that event A does not occur is:
\[ P(A') = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3} \]
- The probability that event B does not occur is:
\[ P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4} \]

3. **Use the formula for the probability of exactly one event occurring:**
- The probability that exactly one of the events A or B occurs can be calculated as:
\[ P(A \text{ and } B') + P(A' \text{ and } B) \]
- This can be expressed using the probabilities:
\[ P(A) \cdot P(B') + P(B) \cdot P(A') \]

4. **Substitute the values into the formula:**
- Substitute \( P(A) \), \( P(B') \), \( P(B) \), and \( P(A') \):
\[ P(A) \cdot P(B') = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12} \]
\[ P(B) \cdot P(A') = \frac{3}{4} \cdot \frac{2}{3} = \frac{6}{12} = \frac{1}{2} \]

5. **Add the two probabilities together:**
- Now, combine the results:
\[ P(A \text{ and } B') + P(A' \text{ and } B) = \frac{1}{12} + \frac{1}{2} \]
- To add these fractions, convert \(\frac{1}{2}\) to have a common denominator of 12:
\[ \frac{1}{2} = \frac{6}{12} \]
- Thus, we have:
\[ \frac{1}{12} + \frac{6}{12} = \frac{7}{12} \]

6. **Final Result:**
- Therefore, the probability that exactly one of the two events A or B occurs is:
\[ \boxed{\frac{7}{12}} \]