Class CBSE Class 12 Mathematics Probability Q #1740
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If $3P(A)=P(B)=\frac{3}{5}$ and $P(A|B)=\frac{2}{3}$ then $P(A\cup B)$ is:
(A) $\frac{3}{5}$
(B) $\frac{1}{5}$
(C) $\frac{2}{5}$
(D) $\frac{2}{15}$
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Detailed Solution

Step 1: Extract individual probabilities

Given the equations $3P(A) = \frac{3}{5}$ and $P(B) = \frac{3}{5}$. From $3P(A) = \frac{3}{5}$, we get $P(A) = \frac{1}{5}$.

Step 2: Find the intersection probability

Using the conditional probability formula $P(A|B) = \frac{P(A \cap B)}{P(B)}$, we substitute the known values: $$\frac{2}{3} = \frac{P(A \cap B)}{3/5}$$ $$P(A \cap B) = \frac{2}{3} \times \frac{3}{5} = \frac{2}{5}$$

Step 3: Calculate the union probability

Using the addition theorem of probability: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ $$P(A \cup B) = \frac{1}{5} + \frac{3}{5} - \frac{2}{5}$$ $$P(A \cup B) = \frac{1+3-2}{5} = \frac{2}{5}$$

Final Answer: 2/5

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to synthesize multiple probability formulas (conditional probability and addition theorem) to solve for an unknown variable.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of algebraic steps and logical operations to derive the final probability value.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to manipulate probability definitions beyond simple textbook definitions.

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