A,B,C are events such that `P(A)=0.3,P(B)=0.4,PC )=0.8,P(A cap B)=0.8,P(A cap C) =0.28, P(A cap B cap C)=0.09`. If `0.75 le P( A cup B cup C) le 1`, then `P(B cap C)` may be

Options

A. 0.21
B. 0.24
C. 0.35
D. 0.5

Answer

To solve the problem, we will use the principle of inclusion-exclusion to calculate \( P(A \cup B \cup C) \) and then find the possible values for \( P(B \cap C) \).

### Step-by-Step Solution:

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

2. **Use the formula for the probability of 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. **Substituting the known values into the formula:**
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.8 - 0.28 - P(B \cap C) + 0.09 \]

4. **Simplifying the expression:**
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.8 - 0.28 + 0.09 - P(B \cap C) \]
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.09 - 0.28 - P(B \cap C) \]
\[ P(A \cup B \cup C) = 0.51 - P(B \cap C) \]

5. **Setting up the inequalities based on the given range for \( P(A \cup B \cup C) \):**
\[ 0.75 \leq P(A \cup B \cup C) \leq 1 \]

6. **Substituting the expression for \( P(A \cup B \cup C) \):**
\[ 0.75 \leq 0.51 - P(B \cap C) \leq 1 \]

7. **Solving the left inequality:**
\[ 0.75 \leq 0.51 - P(B \cap C) \]
\[ P(B \cap C) \leq 0.51 - 0.75 \]
\[ P(B \cap C) \leq -0.24 \quad \text{(not possible, discard)} \]

8. **Solving the right inequality:**
\[ 0.51 - P(B \cap C) \leq 1 \]
\[ -P(B \cap C) \leq 1 - 0.51 \]
\[ -P(B \cap C) \leq 0.49 \]
\[ P(B \cap C) \geq -0.49 \quad \text{(not useful)} \]

9. **Combining both inequalities:**
Since \( P(B \cap C) \) must be a non-negative probability, we can conclude:
\[ 0 \leq P(B \cap C) \leq 0.48 \]

### Conclusion:
Thus, the possible values for \( P(B \cap C) \) can be any value in the range \( [0, 0.48] \).