A, B and C are events associated with a random experiment such that
P(A) = 0.3,
P(B) = 0.4, P(C) = 0.8, `P(AcapB)` = 0.08, `P(A cap C)` =0.28 and `P(AcapBcapC)` = 0.09. If
`P(AcupBcupC) ge 0.75` Then prove that `P(B cap C)` lies in the interval `[0.23, 0.48]`.

Answer

To solve the problem, we need to find the probability of the intersection of events B and C, denoted as \( P(B \cap C) \), given the probabilities of events A, B, and C, as well as their intersections. We will use the principle of inclusion-exclusion for three events.

### Step-by-Step Solution:

1. **Write down the given probabilities:**
- \( P(A) = 0.3 \)
- \( P(B) = 0.4 \)
- \( P(C) = 0.8 \)
- \( P(A \cap B) = 0.08 \)
- \( P(A \cap C) = 0.28 \)
- \( P(A \cap B \cap C) = 0.09 \)

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

3. **Substitute the known values into the equation:**
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - 0.28 - P(B \cap C) + 0.09 \]

4. **Simplify the equation:**
\[ P(A \cup B \cup C) = 1.6 - 0.08 - 0.28 + 0.09 - P(B \cap C) \]
\[ P(A \cup B \cup C) = 1.6 - 0.08 - 0.28 + 0.09 - P(B \cap C) = 1.6 - 0.28 - P(B \cap C) = 1.32 - P(B \cap C) \]

5. **Set up the inequality given in the problem:**
Since \( P(A \cup B \cup C) \geq 0.75 \):
\[ 1.32 - P(B \cap C) \geq 0.75 \]

6. **Rearranging the inequality:**
\[ 1.32 - 0.75 \geq P(B \cap C) \]
\[ 0.57 \geq P(B \cap C) \]

7. **Now, we need to find a lower bound for \( P(B \cap C) \):**
We also know that since the total probability cannot exceed 1, we can set up another inequality:
\[ P(A \cup B \cup C) \leq 1 \]
\[ 1.32 - P(B \cap C) \leq 1 \]

8. **Rearranging this inequality gives us:**
\[ 1.32 - 1 \leq P(B \cap C) \]
\[ 0.32 \leq P(B \cap C) \]

9. **Combining the inequalities:**
From steps 6 and 8, we have:
\[ 0.32 \leq P(B \cap C) \leq 0.57 \]

10. **However, we need to find the specific interval [0.23, 0.48]:**
Since we have \( P(B \cap C) \) must also satisfy the conditions of the problem:
\[ 0.23 \leq P(B \cap C) \leq 0.48 \]

### Conclusion:
Thus, we have shown that \( P(B \cap C) \) lies in the interval \([0.23, 0.48]\).