Let A and B be two events such that `P(A cup B) ge 3//4 and 1//8 le P(A cap B) le 3//8`
Statement 1 : `P(A) +P(B) ge 7//8`
Statement 2 : `P(A) +P(B) le 11//8`

Options

A. Statement 1 and statement 2 are both false
B. Statement 1 and statement 2 are both true
C. Statement 1 is true and statement 2 is false
D. Statement 1 is false and statement 2 is true

Answer

To solve the problem, we need to analyze the given probabilities and derive the required inequalities for \( P(A) + P(B) \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know:
- \( P(A \cup B) \geq \frac{3}{4} \)
- \( \frac{1}{8} \leq P(A \cap B) \leq \frac{3}{8} \)

2. **Using the Formula for Union of Two Events**:
The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
From this, we can express \( P(A) + P(B) \):
\[ P(A) + P(B) = P(A \cup B) + P(A \cap B) \]

3. **Substituting the Bounds**:
We substitute the bounds for \( P(A \cup B) \) and \( P(A \cap B) \):
- Lower Bound:
\[ P(A) + P(B) \geq \frac{3}{4} + \frac{1}{8} \]
To add these fractions, convert \( \frac{3}{4} \) to eighths:
\[ \frac{3}{4} = \frac{6}{8} \implies P(A) + P(B) \geq \frac{6}{8} + \frac{1}{8} = \frac{7}{8} \]

- Upper Bound:
\[ P(A) + P(B) \leq 1 + \frac{3}{8} \]
Again, convert \( 1 \) to eighths:
\[ 1 = \frac{8}{8} \implies P(A) + P(B) \leq \frac{8}{8} + \frac{3}{8} = \frac{11}{8} \]

4. **Conclusion**:
From the calculations, we have:
\[ \frac{7}{8} \leq P(A) + P(B) \leq \frac{11}{8} \]
This means:
- Statement 1: \( P(A) + P(B) \geq \frac{7}{8} \) is **True**.
- Statement 2: \( P(A) + P(B) \leq \frac{11}{8} \) is **True**.

### Final Result:
Both statements are true.

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