A,B,C are three events for which `P(A)=0.4,P(B)=0.6,P(C ) =0.5 , P( A cup B)=0.75, P(A cap C) =0.35` and `P(A cap B cap C)= 0.2` if `P(A cup B cup C) le 0.75`, then `P(B cap C)` can take values.

Options

A. 0.1
B. 0.2
C. 0.3
D. 0.5

Answer

To solve the problem, we need to find the possible values of \( P(B \cap C) \) given the probabilities of events A, B, and C, as well as their intersections and unions. Let's break it down step by step.

### Step 1: Identify Given Values
We have the following probabilities:
- \( P(A) = 0.4 \)
- \( P(B) = 0.6 \)
- \( P(C) = 0.5 \)
- \( P(A \cup B) = 0.75 \)
- \( P(A \cap C) = 0.35 \)
- \( P(A \cap B \cap C) = 0.2 \)

### Step 2: Calculate \( P(A \cap B) \)
Using the formula for the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ 0.75 = 0.4 + 0.6 - P(A \cap B) \]
\[ 0.75 = 1.0 - P(A \cap B) \]
\[ P(A \cap B) = 1.0 - 0.75 = 0.25 \]

### Step 3: Use the Formula for \( P(A \cup B \cup C) \)
The formula for the union of three events is:
\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \]
Substituting the known values:
\[ P(A \cup B \cup C) \leq 0.75 \]
\[ P(A \cup B \cup C) = 0.4 + 0.6 + 0.5 - 0.25 - P(B \cap C) - 0.35 + 0.2 \]
\[ P(A \cup B \cup C) = 1.7 - P(B \cap C) \]

### Step 4: Set Up the Inequality
Since \( P(A \cup B \cup C) \leq 0.75 \):
\[ 1.7 - P(B \cap C) \leq 0.75 \]
Rearranging gives:
\[ -P(B \cap C) \leq 0.75 - 1.7 \]
\[ -P(B \cap C) \leq -0.95 \]
Multiplying through by -1 (which reverses the inequality):
\[ P(B \cap C) \geq 0.95 \]

### Step 5: Find the Upper Bound for \( P(B \cap C) \)
Since \( P(B \cap C) \) cannot exceed the individual probabilities of B and C:
\[ P(B \cap C) \leq \min(P(B), P(C)) = \min(0.6, 0.5) = 0.5 \]

### Step 6: Combine Results
From the inequalities we derived:
\[ 0.35 \leq P(B \cap C) \leq 0.5 \]

### Conclusion
Thus, the possible values for \( P(B \cap C) \) can range from 0.35 to 0.5.