Class CBSE Class 12 Mathematics Probability Q #691
COMPETENCY BASED
APPLY
1 Marks 2025 AISSCE(Board Exam) MCQ SINGLE
If \(P(A)=\frac{1}{7}\), \(P(B)=\frac{5}{7}\) and \(P(A\cap B)=\frac{4}{7},\) then \(P(\overline{A}|B)\) is:
(A) \(\frac{6}{7}\)
(B) \(\frac{3}{4}\)
(C) \(\frac{4}{5}\)
(D) \(\frac{1}{5}\)
Correct Answer: D
Explanation
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Step-by-Step Solution

We are given \(P(A) = \frac{1}{7}\), \(P(B) = \frac{5}{7}\), and \(P(A \cap B) = \frac{4}{7}\). We need to find \(P(\overline{A}|B)\).

By definition of conditional probability, \(P(\overline{A}|B) = \frac{P(\overline{A} \cap B)}{P(B)}\).

We know that \(P(B) = P(A \cap B) + P(\overline{A} \cap B)\). Therefore, \(P(\overline{A} \cap B) = P(B) - P(A \cap B)\).

Substituting the given values, we have \(P(\overline{A} \cap B) = \frac{5}{7} - \frac{4}{7} = \frac{1}{7}\).

Now, we can find \(P(\overline{A}|B) = \frac{P(\overline{A} \cap B)}{P(B)} = \frac{\frac{1}{7}}{\frac{5}{7}} = \frac{1}{5}\).

Correct Answer: \(\frac{1}{5}\)

AI Suggestion: Option D

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the formula for conditional probability and the complement of an event to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to follow a specific procedure to calculate the conditional probability, involving the use of formulas and algebraic manipulation.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply the concepts of probability and conditional probability in a problem-solving context, rather than simply recalling definitions or theorems.