CBSE Class 12 Mathematics Probability Q #1801
COMPETENCY BASED
APPLY
3 Marks 2026 AISSCE(Board Exam) SA
Mother, Father and Son line up at random for a family picture. Let events E: Son on one end and F: Father in the middle. Find $P(E/F)$.
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Detailed Solution

Step 1: Define the Sample Space

Let the family members be M (Mother), F (Father), and S (Son). The total number of ways to arrange 3 people in a line is $3! = 6$. The sample space $S$ is: {MFS, MSF, FMS, FSM, SMF, SFM}. Thus, $n(S) = 6$.

Step 2: Identify Event F

Event F is defined as 'Father in the middle'. The arrangements satisfying this are {MFS, SFM}. Thus, $n(F) = 2$. The probability $P(F) = \frac{2}{6} = \frac{1}{3}$.

Step 3: Identify Event E and Intersection

Event E is 'Son on one end'. The intersection $E \cap F$ represents 'Father in the middle AND Son on one end'. Looking at the set F = {MFS, SFM}, both outcomes satisfy the condition of the Son being at one end. Thus, $E \cap F = \{MFS, SFM\}$, and $n(E \cap F) = 2$.

Step 4: Calculate Conditional Probability

Using the formula $P(E/F) = \frac{P(E \cap F)}{P(F)}$, we substitute the values: $$P(E/F) = \frac{2/6}{2/6} = 1$$

Final Answer: 1

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to translate a verbal description of a probability experiment into a sample space and apply the conditional probability formula.
Knowledge Dimension: PROCEDURAL
Justification: The student must follow a specific sequence of steps: defining the sample space, identifying subsets, and executing the conditional probability algorithm.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. This question tests the student's ability to interpret events in a constrained sample space, which is a core competency in the Probability chapter.

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