Class CBSE Class 12 Mathematics Probability Q #1804
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
A die is rolled. Consider events: $A=\{1,2,5\}$, $B=\{3,5\}$, $C=\{2,3,4,5\}$ and hence find: (i) $P(A|C)$ and $P(C|A)$ (ii) $P(A\cap B|C)$ and $P(A\cup B|C)$.

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Detailed Solution

Step 1: Define Sample Space and Probabilities

For a single die roll, the sample space is $S = \{1, 2, 3, 4, 5, 6\}$. The total number of outcomes is $n(S) = 6$. The probability of any single outcome is $1/6$.

Step 2: Calculate Conditional Probabilities for (i)

We use the formula $P(E|F) = \frac{P(E \cap F)}{P(F)}$. Given $A=\{1,2,5\}$ and $C=\{2,3,4,5\}$, the intersection $A \cap C = \{2, 5\}$. Thus, $n(A \cap C) = 2$.

$$P(A|C) = \frac{n(A \cap C)}{n(C)} = \frac{2}{4} = \frac{1}{2}$$

$$P(C|A) = \frac{n(C \cap A)}{n(A)} = \frac{2}{3}$$

Step 3: Calculate Conditional Probabilities for (ii)

First, find $A \cap B = \{5\}$ and $A \cup B = \{1, 2, 3, 5\}$.

For $P(A \cap B | C)$: The intersection $(A \cap B) \cap C = \{5\}$. Thus, $n((A \cap B) \cap C) = 1$.

$$P(A \cap B | C) = \frac{1}{4}$$

For $P(A \cup B | C)$: The intersection $(A \cup B) \cap C = \{2, 3, 5\}$. Thus, $n((A \cup B) \cap C) = 3$.

$$P(A \cup B | C) = \frac{3}{4}$$

Final Answer: (i) P(A|C)=1/2, P(C|A)=2/3; (ii) P(A∩B|C)=1/4, P(A∪B|C)=3/4

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to utilize the conditional probability formula in a specific set-theoretic context.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a sequence of steps: identifying intersections, calculating cardinalities, and applying the conditional probability ratio.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. It tests the student's ability to manipulate sets within the framework of probability theory, which is a core competency in the 'Probability' unit.