A and B are any to events such that `P(A)=0.3, P(B)=0.8 and P(A cap B) =0.16`, find the probability that exactly one of the events happens.

Answer

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

### Step-by-Step Solution:

1. **Identify Given Probabilities**:
- \( P(A) = 0.3 \)
- \( P(B) = 0.8 \)
- \( P(A \cap B) = 0.16 \)

2. **Calculate the Probability of Only A**:
- The probability of only A occurring (not B) can be calculated as:
\[ P(\text{Only A}) = P(A) - P(A \cap B) \]
- Substituting the values:
\[ P(\text{Only A}) = 0.3 - 0.16 = 0.14 \]

3. **Calculate the Probability of Only B**:
- The probability of only B occurring (not A) can be calculated as:
\[ P(\text{Only B}) = P(B) - P(A \cap B) \]
- Substituting the values:
\[ P(\text{Only B}) = 0.8 - 0.16 = 0.64 \]

4. **Calculate the Probability of Exactly One Event Occurring**:
- The probability that exactly one of the events occurs (either A or B, but not both) is the sum of the probabilities of only A and only B:
\[ P(\text{Exactly one of A or B}) = P(\text{Only A}) + P(\text{Only B}) \]
- Substituting the values:
\[ P(\text{Exactly one of A or B}) = 0.14 + 0.64 = 0.78 \]

5. **Final Result**:
- The probability that exactly one of the events happens is:
\[ P(\text{Exactly one of A or B}) = 0.78 \]
- This can also be expressed as a fraction:
\[ 0.78 = \frac{78}{100} = \frac{39}{50} \]

### Conclusion:
The required probability that exactly one of the events A or B occurs is \( \frac{39}{50} \).