The probability that at least one of the events A and B occurs is `0.6`. If A and B occur simultaneously with probability `0.2`, find `P(barA)+P(barB)`

Answer

To solve the problem, we need to find \( P(\bar{A}) + P(\bar{B}) \) given that \( P(A \cup B) = 0.6 \) and \( P(A \cap B) = 0.2 \).

### Step-by-step Solution:

1. **Identify the given probabilities**:
- \( P(A \cup B) = 0.6 \) (the probability that at least one of the events A or B occurs)
- \( P(A \cap B) = 0.2 \) (the probability that both events A and B occur simultaneously)

2. **Use the formula for the union of two events**:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ 0.6 = P(A) + P(B) - 0.2 \]

3. **Rearrange the equation to find \( P(A) + P(B) \)**:
\[ P(A) + P(B) = 0.6 + 0.2 = 0.8 \]

4. **Express \( P(A) \) and \( P(B) \) in terms of their complements**:
We know that:
\[ P(A) + P(\bar{A}) = 1 \quad \text{and} \quad P(B) + P(\bar{B}) = 1 \]
Therefore, we can express \( P(A) \) and \( P(B) \) as:
\[ P(A) = 1 - P(\bar{A}) \quad \text{and} \quad P(B) = 1 - P(\bar{B}) \]

5. **Substitute these expressions into the equation**:
\[ (1 - P(\bar{A})) + (1 - P(\bar{B})) = 0.8 \]
Simplifying this gives:
\[ 2 - P(\bar{A}) - P(\bar{B}) = 0.8 \]

6. **Rearranging to find \( P(\bar{A}) + P(\bar{B}) \)**:
\[ P(\bar{A}) + P(\bar{B}) = 2 - 0.8 = 1.2 \]

### Final Answer:
\[ P(\bar{A}) + P(\bar{B}) = 1.2 \]