Question

Which is the following statement is true?

• A. \(P(C_1 \cap C_2) \ge P(C_1) + P(C_2)\)
• B. \(P(C_1 \cup C_2) \ge P(C_1) + P(C_2)\)
• C. \(P(C_1 \cap C_2) \ge P(C_1) + P(C_2) - 1\)
• D. \(P(C_1 \cap C_2) \le P(C_1) + P(C_2) - 1\)

Hint:

Bonferroni's inequality is a direct consequence of the fact that the probability of the union of any events cannot exceed 1:
\[ P(A \cup B) \le 1 \implies P(A \cap B) \ge P(A) + P(B) - 1 \]
This is a helpful identity to memorize.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This question asks about Bonferroni's inequality for two events, which relates the joint probability of two events to their individual probabilities.

Step 2: Key Formula or Approach:
The addition theorem of probability states that:
\[ P(C_1 \cup C_2) = P(C_1) + P(C_2) - P(C_1 \cap C_2) \]
Since any probability value must be less than or equal to 1, we can write:
\[ P(C_1 \cup C_2) \le 1 \]

Step 3: Detailed Explanation:
Let us use the addition theorem and the fact that probability is bounded by 1 to derive the relationship:
Substitute the addition theorem into our inequality:
\[ P(C_1) + P(C_2) - P(C_1 \cap C_2) \le 1 \]
Now, rearrange the terms to isolate the joint probability, \(P(C_1 \cap C_2)\).
Add \(P(C_1 \cap C_2)\) to both sides of the inequality:
\[ P(C_1) + P(C_2) \le 1 + P(C_1 \cap C_2) \]
Subtract 1 from both sides:
\[ P(C_1) + P(C_2) - 1 \le P(C_1 \cap C_2) \]
We can rewrite this inequality as:
\[ P(C_1 \cap C_2) \ge P(C_1) + P(C_2) - 1 \]
This inequality is known as Bonferroni's inequality for two events.
It provides a lower bound for the joint probability of two events when their individual probabilities are known.
This matches the third option.

Step 4: Final Answer:
Therefore, the correct option is (C).