If A, B, C are events such that `P(A)=0.3,P(B)=0.4,P(C )=0.8`
`P(AnnB)=0.08,P(AnnC)=0.28`
`P(AnnBnnC)=0.09`
If `P(AuuBuuC)ge0.75`, then find the range of `x=P(BnnC)` lies in the interval.

Options

A. `0.23lexle0.48`
B. `0.23lexle0.47`
C. `0.22lexle0.48`
D. None of these

Answer

To solve the problem, we will use the principle of inclusion-exclusion for three events A, B, and C. We are given the probabilities of individual events and their intersections, and we need to find the range of \( P(B \cap C) \) given that \( P(A \cup B \cup C) \geq 0.75 \).

### Step-by-Step Solution:

1. **Identify Given Values:**
- \( P(A) = 0.3 \)
- \( P(B) = 0.4 \)
- \( P(C) = 0.8 \)
- \( P(A \cap B) = 0.08 \)
- \( P(A \cap C) = 0.28 \)
- \( P(A \cap B \cap C) = 0.09 \)
- Let \( P(B \cap C) = x \)

2. **Use the Inclusion-Exclusion Principle:**
The formula for the union of three events is:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \]
Substituting the known values:
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - 0.28 - x + 0.09 \]

3. **Simplify the Equation:**
Combine the constants:
\[ P(A \cup B \cup C) = 1.6 - 0.08 - 0.28 + 0.09 - x \]
\[ P(A \cup B \cup C) = 1.6 - 0.08 - 0.28 + 0.09 - x = 1.6 - 0.27 - x = 1.33 - x \]

4. **Set Up the Inequality:**
We know that:
\[ P(A \cup B \cup C) \geq 0.75 \]
Therefore:
\[ 1.33 - x \geq 0.75 \]

5. **Solve for \( x \):**
Rearranging the inequality:
\[ 1.33 - 0.75 \geq x \]
\[ 0.58 \geq x \]
This gives us the upper limit for \( x \):
\[ x \leq 0.58 \]

6. **Find the Lower Limit for \( x \):**
We also know that \( P(B \cap C) \) cannot be negative, so:
\[ x \geq 0 \]

7. **Combine the Results:**
Thus, the range for \( P(B \cap C) \) is:
\[ 0 \leq x \leq 0.58 \]

### Final Answer:
The range of \( P(B \cap C) \) lies in the interval \( [0, 0.58] \).