If A and B are arbitrary events then

Options

A. `P(A capB)leP(A)+P(B)`
B. `P(A cupB)leP(A)+P(B)`
C. `P(A cupB)=P(A)+P(B)`
D. `P(A cupB) geP(A)+P(B)`

Answer

To solve the problem, we need to analyze the relationship between the probabilities of two arbitrary events A and B. We will derive the inequality that relates the probabilities of these events.

### Step-by-Step Solution:

1. **Understanding the Union of Events**:
The probability of the union of two events A and B is given by the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

2. **Setting Up the Non-Negative Condition**:
Since probabilities cannot be negative, we need to ensure that the expression for the probability of the union is non-negative:
\[ P(A \cup B) \geq 0 \]
This leads us to:
\[ P(A) + P(B) - P(A \cap B) \geq 0 \]

3. **Rearranging the Inequality**:
Rearranging the above inequality gives:
\[ P(A) + P(B) \geq P(A \cap B) \]

4. **Conclusion**:
This shows that the probability of the intersection of events A and B is less than or equal to the sum of their individual probabilities:
\[ P(A \cap B) \leq P(A) + P(B) \]

Thus, we have established that for any two arbitrary events A and B, the inequality holds true.

### Final Answer:
The correct option is:
\[ P(A \cap B) \leq P(A) + P(B) \]