Question

If $C_1$ and $C_2$ are subset of sample space S. Then which is correct:

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

Hint:

To visualize these set relationships quickly, use a Venn Diagram:
- The overlapping region (Intersection) is always smaller than or equal to either individual circle.
- Both circles combined (Union) always cover an area larger than or equal to the overlapping region.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Set theory and probability axioms define the relationships among different events (subsets) within a sample space (\(S\)).
The probability of a subset is always proportional to its size relative to the sample space, meaning larger sets have higher probabilities.

Step 2: Key Formula or Approach:
For any two events \(C_1\) and \(C_2\) in a sample space \(S\):
- Intersection (\(C_1 \cap C_2\)) represents the elements common to both sets.
- Union (\(C_1 \cup C_2\)) represents all elements contained in either set.
The set inclusion relationship dictates:
\[ (C_1 \cap C_2) \subseteq C_1 \subseteq (C_1 \cup C_2) \] \[ (C_1 \cap C_2) \subseteq C_2 \subseteq (C_1 \cup C_2) \] According to the monotonicity property of probability, if \(A \subseteq B\), then:
\[ P(A) \le P(B) \]

Step 3: Detailed Explanation:
Let us evaluate each of the given inequalities:
- Evaluating Option (A) and (B): Since \((C_1 \cap C_2) \subseteq C_1\), it must be true that:
\[ P(C_1 \cap C_2) \le P(C_1) \] Therefore, the inverse relationships shown in (A) and (B) are incorrect.
- Evaluating Option (C): Since the intersection is a subset of the union, the probability of the intersection must be less than or equal to the probability of the union:
\[ P(C_1 \cap C_2) \le P(C_1 \cup C_2) \] This inequality is always true, regardless of whether the events are independent, dependent, or mutually exclusive.
- Evaluating Option (D): According to the additive law of probability:
\[ P(C_1 \cup C_2) = P(C_1) + P(C_2) - P(C_1 \cap C_2) \] Since \(P(C_1 \cap C_2) \ge 0\), the union probability must be less than or equal to the sum of the individual probabilities:
\[ P(C_1 \cup C_2) \le P(C_1) + P(C_2) \] Therefore, Option (D) is incorrect.

Step 4: Final Answer:
The mathematically correct statement is \(P(C_1 \cap C_2) \le P(C_1 \cup C_2)\), matching Option (C).