If A and B are two events such that P(A) = `(1)/(2)` and P(B) = `(2)/(3)`, then show that
(a) `P(Auu B) ge (2)/(3) (b) (1)/(6) le P(A nn B) le (1)/(2)`
(c ) `P(A nn barB) le (1)/(3) (d) (1)/(6) le P(barAnn B) le (1)/(2)`

Options

A. `P(A cup B) ge (2)/(3)`
B. `P(A cap barB) le (1)/(3)`
C. `(1)/(6) le P(A cap B) le (1)/(2)`
D. `(1)/(6) le P(barA cap B) le (1)/(2)`

Answer

To solve the problem step-by-step, we will analyze each part of the question based on the given probabilities of events A and B.

### Given:
- \( P(A) = \frac{1}{2} \)
- \( P(B) = \frac{2}{3} \)

### Part (a): Show that \( P(A \cup B) \geq \frac{2}{3} \)

1. **Use the formula for the union of two events:**
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

2. **Substituting the known values:**
\[ P(A \cup B) = \frac{1}{2} + \frac{2}{3} - P(A \cap B) \]

3. **Finding a common denominator to add \( P(A) \) and \( P(B) \):**
\[ P(A) + P(B) = \frac{3}{6} + \frac{4}{6} = \frac{7}{6} \]

4. **Thus, we have:**
\[ P(A \cup B) = \frac{7}{6} - P(A \cap B) \]

5. **Since probabilities cannot be negative, we have:**
\[ P(A \cap B) \geq 0 \]

6. **This leads to:**
\[ P(A \cup B) \geq \frac{7}{6} - 0 = \frac{7}{6} \]

7. **Since \( \frac{7}{6} > \frac{2}{3} \), we conclude:**
\[ P(A \cup B) \geq \frac{2}{3} \]