Assertion (A) : If `A sub B and B sub A ` then P(A) = P(B)
Reason (R) : If `A sub B` then ` P (bar(A)) le P (bar (B))`

Options

A. Both A and R are true and R is the correct explanation of a
B. Both A and R are true but R is NOT the correct explanation of A
C. A is true but is false
D. A is false but R is true

Answer

To solve the given question, we need to analyze both the assertion (A) and the reason (R) provided.

### Step 1: Analyze the Assertion (A)
The assertion states: "If A ⊆ B and B ⊆ A, then P(A) = P(B)."

- If A is a subset of B (A ⊆ B) and B is a subset of A (B ⊆ A), this implies that A and B contain exactly the same elements. Therefore, we can conclude that A = B.
- Since A and B are equal sets, their probabilities must also be equal. Thus, P(A) = P(B).

**Conclusion for Assertion (A)**: True.

### Step 2: Analyze the Reason (R)
The reason states: "If A ⊆ B, then P(bar(A)) ≤ P(bar(B))."

- Here, "bar(A)" refers to the complement of A, which is the set of outcomes not in A.
- The theorem states that if A is a subset of B (A ⊆ B), then the probability of A is less than or equal to the probability of B, i.e., P(A) ≤ P(B).
- The complement rule states that P(bar(A)) = 1 - P(A) and P(bar(B)) = 1 - P(B).
- If A ⊆ B, then it follows that P(A) ≤ P(B), which implies that P(bar(A)) ≥ P(bar(B)). Therefore, the statement P(bar(A)) ≤ P(bar(B)) is incorrect.

**Conclusion for Reason (R)**: False.

### Final Conclusion
- Assertion (A) is true: P(A) = P(B) when A ⊆ B and B ⊆ A.
- Reason (R) is false: The correct statement is P(bar(A)) ≥ P(bar(B)) if A ⊆ B.

### Summary
- Assertion (A): True
- Reason (R): False

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