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 step by step, we will use the formula for the probability of the union of three events and the given probabilities.

### Step 1: Write down the given probabilities
We have the following probabilities:
- \( P(A) = 0.3 \)
- \( P(B) = 0.4 \)
- \( P(C) = 0.8 \)
- \( P(A \cap B) = 0.08 \)
- \( P(A \cap C) = 0.28 \)
- \( P(A \cap B \cap C) = 0.09 \)

### Step 2: Use the formula for the union of three events
The formula for the probability of the union of three events \( A \), \( B \), and \( C \) is given by:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \]

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula, we get:
\[ P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - P(B \cap C) - 0.28 + 0.09 \]

### Step 4: Simplify the equation
Now, simplify the equation:
\[ P(A \cup B \cup C) = 1.6 - P(B \cap C) - 0.08 - 0.28 + 0.09 \]
\[ P(A \cup B \cup C) = 1.6 - P(B \cap C) - 0.27 \]
\[ P(A \cup B \cup C) = 1.33 - P(B \cap C) \]

### Step 5: Set up inequalities based on the given conditions
We know that:
\[ 0.75 \leq P(A \cup B \cup C) \leq 1 \]
Substituting our expression for \( P(A \cup B \cup C) \):
\[ 0.75 \leq 1.33 - P(B \cap C) \leq 1 \]

### Step 6: Solve the inequalities
1. For the left inequality:
\[ 0.75 \leq 1.33 - P(B \cap C) \]
Rearranging gives:
\[ P(B \cap C) \leq 1.33 - 0.75 \]
\[ P(B \cap C) \leq 0.58 \]

2. For the right inequality:
\[ 1.33 - P(B \cap C) \leq 1 \]
Rearranging gives:
\[ P(B \cap C) \geq 1.33 - 1 \]
\[ P(B \cap C) \geq 0.33 \]

### Step 7: Combine the results
From the inequalities, we have:
\[ 0.33 \leq P(B \cap C) \leq 0.58 \]

### Conclusion
Thus, \( P(B \cap C) \) may be any value in the range \( [0.33, 0.58] \).