CBSE Class 12 Mathematics Probability Q #1738
COMPETENCY BASED
APPLY
1 Marks 2026 AISSCE(Board Exam) MCQ SINGLE
If E and F are two independent events such that $P(E)=\frac{3}{10}$, $P(E\cup F)=\frac{1}{2}$ then $P(E|F)-P(F|E)$ is equal to:
(A) $\frac{2}{7}$
(B) $\frac{3}{35}$
(C) $\frac{1}{70}$
(D) $\frac{1}{7}$
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Correct Answer: C

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Detailed Solution

Step 1: Find P(F) using the Addition Theorem

Given $P(E) = \frac{3}{10}$ and $P(E \cup F) = \frac{1}{2}$. Since $E$ and $F$ are independent, $P(E \cap F) = P(E) \cdot P(F) = \frac{3}{10}P(F)$. Using the formula $P(E \cup F) = P(E) + P(F) - P(E \cap F)$:

$$\frac{1}{2} = \frac{3}{10} + P(F) - \frac{3}{10}P(F)$$ $$\frac{1}{2} - \frac{3}{10} = P(F)(1 - \frac{3}{10})$$ $$\frac{2}{10} = P(F)(\frac{7}{10}) \implies P(F) = \frac{2}{7}$$

Step 2: Calculate Conditional Probabilities

For independent events, $P(E|F) = P(E)$ and $P(F|E) = P(F)$.

$$P(E|F) = \frac{3}{10}$$ $$P(F|E) = \frac{2}{7}$$

Step 3: Compute the final difference

Subtract the values obtained:

$$P(E|F) - P(F|E) = \frac{3}{10} - \frac{2}{7}$$ $$= \frac{21 - 20}{70} = \frac{1}{70}$$

Final Answer: 1/70

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the properties of independent events and the addition theorem of probability to solve for unknown variables.
Knowledge Dimension: PROCEDURAL
Justification: The problem requires a sequential execution of algebraic steps and probability definitions to reach the solution.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the conceptual understanding of independence in probability, which is a core component of the Probability chapter in the NCERT curriculum.

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