Question:
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B){P(B)} \).}
Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B){P(B)} \).}
Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B){P(B)} \).}
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Tip 1: In conditional probability, the given condition reduces the size of the sample space.
Tip 2: Remember that \( P(A|B) \) involves the intersection \( A \cap B \), not the union \( A \cup B \).
Updated On: Sep 10, 2026
- Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true and Reason (R) is false.
- Assertion (A) is false and Reason (R) is true.
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The Correct Option is C
Solution and Explanation
Concept:
• Conditional Probability: The probability of an event \( A \) given that event \( B \) has occurred is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(B) > 0 \).
• For an unbiased die, the sample space is \( S = \{1, 2, 3, 4, 5, 6\} \).
Step 1: Verify the Assertion (A)
Let \( A \) be the event of getting a prime number. \[ A = \{2, 3, 5\} \implies n(A) = 3 \] Let \( B \) be the event that the number appearing on the die is odd. \[ B = \{1, 3, 5\} \implies n(B) = 3 \] The intersection \( A \cap B \) is the set of numbers that are both prime and odd: \[ A \cap B = \{3, 5\} \implies n(A \cap B) = 2 \] Calculating the conditional probability: \[ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{3} \] Thus, Assertion (A) is True.
Step 2: Verify the Reason (R)
The formula provided in the Reason is: \[ P(A|B) = \frac{P(A \cup B)}{P(B)} \] This formula is incorrect. The correct formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 \] Therefore, Reason (R) is False.
Step 3: Check the relationship between A and R
Assertion (A) is True, as \[ P(A|B) = \frac{2}{3}. \] However, Reason (R) is False because conditional probability is calculated using the intersection \( A \cap B \), not the union \( A \cup B \). Therefore, Assertion (A) is true and Reason (R) is false.
• Conditional Probability: The probability of an event \( A \) given that event \( B \) has occurred is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( P(B) > 0 \).
• For an unbiased die, the sample space is \( S = \{1, 2, 3, 4, 5, 6\} \).
Step 1: Verify the Assertion (A)
Let \( A \) be the event of getting a prime number. \[ A = \{2, 3, 5\} \implies n(A) = 3 \] Let \( B \) be the event that the number appearing on the die is odd. \[ B = \{1, 3, 5\} \implies n(B) = 3 \] The intersection \( A \cap B \) is the set of numbers that are both prime and odd: \[ A \cap B = \{3, 5\} \implies n(A \cap B) = 2 \] Calculating the conditional probability: \[ P(A|B) = \frac{n(A \cap B)}{n(B)} = \frac{2}{3} \] Thus, Assertion (A) is True.
Step 2: Verify the Reason (R)
The formula provided in the Reason is: \[ P(A|B) = \frac{P(A \cup B)}{P(B)} \] This formula is incorrect. The correct formula for conditional probability is: \[ P(A|B) = \frac{P(A \cap B)}{P(B)}, \quad P(B) > 0 \] Therefore, Reason (R) is False.
Step 3: Check the relationship between A and R
Assertion (A) is True, as \[ P(A|B) = \frac{2}{3}. \] However, Reason (R) is False because conditional probability is calculated using the intersection \( A \cap B \), not the union \( A \cup B \). Therefore, Assertion (A) is true and Reason (R) is false.
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