`A` and `B` are `2` events such that `P(A)=(3)/(4)` and `P(B)=(5)/(8)`. If `x` and `y` are the possible minimum and maximum values of `P(AnnB)`, then the value of `a+b` is

Options

A. `0.5`
B. `0.8`
C. `0.9`
D. `1`

Explanation

To solve the problem, we need to find the minimum and maximum values of \( P(A \cap B) \) given \( P(A) = \frac{3}{4} \) and \( P(B) = \frac{5}{8} \).

### Step-by-Step Solution:

1. **Understanding the Events**:
- We have two events \( A \) and \( B \) with probabilities \( P(A) = \frac{3}{4} \) and \( P(B) = \frac{5}{8} \).

2. **Finding the Maximum Value of \( P(A \cap B) \)**:
- The maximum value of \( P(A \cap B) \) cannot exceed the probability of either event. Therefore:
\[ P(A \cap B) \leq P(B) \]
- Thus, the maximum value is:
\[ P(A \cap B) \leq \frac{5}{8} \]

3. **Finding the Minimum Value of \( P(A \cap B) \)**:
- We can use the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Since the probability of any event cannot exceed 1, we have:
\[ P(A \cup B) \leq 1 \]
- Substituting the known values:
\[ P(A) + P(B) - P(A \cap B) \leq 1 \]
- Plugging in the values:
\[ \frac{3}{4} + \frac{5}{8} - P(A \cap B) \leq 1 \]
- To combine the fractions, we find a common denominator (which is 8):
\[ \frac{3}{4} = \frac{6}{8} \]
- Thus:
\[ \frac{6}{8} + \frac{5}{8} - P(A \cap B) \leq 1 \]
- This simplifies to:
\[ \frac{11}{8} - P(A \cap B) \leq 1 \]
- Rearranging gives:
\[ P(A \cap B) \geq \frac{11}{8} - 1 = \frac{11}{8} - \frac{8}{8} = \frac{3}{8} \]

4. **Conclusion**:
- We have found that:
\[ \frac{3}{8} \leq P(A \cap B) \leq \frac{5}{8} \]
- Thus, the minimum value \( x = \frac{3}{8} \) and the maximum value \( y = \frac{5}{8} \).

5. **Calculating \( a + b \)**:
- Here, \( a = \frac{3}{8} \) and \( b = \frac{5}{8} \).
- Therefore:
\[ a + b = \frac{3}{8} + \frac{5}{8} = \frac{8}{8} = 1 \]

### Final Answer:
The value of \( a + b \) is \( 1 \).