Class CBSE Class 12 Mathematics Probability Q #1368
KNOWLEDGE BASED
REMEMBER
3 Marks 2025 AISSCE(Board Exam) SA
The probability distribution for the number of students being absent in a class on a Saturday is as follows: X: 0, 2, 4, 5; $P(X)$: p, 2p, 3p, p. Where X is the number of students absent. (i) Calculate p. (ii) Calculate the mean of the number of absent students on Saturday.
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Detailed Solution

Step 1: Calculate p

The sum of all probabilities in a probability distribution must equal 1. Therefore, we have: $p + 2p + 3p + p = 1$

Step 2: Solve for p

Combining the terms, we get: $7p = 1$ Dividing both sides by 7, we find: $p = \frac{1}{7}$

Step 3: Calculate the mean

The mean (or expected value) of a discrete random variable X is given by: $E(X) = \sum [X \cdot P(X)]$

Step 4: Substitute the values

Substituting the given values and the calculated value of $p$, we have: $E(X) = (0 \cdot p) + (2 \cdot 2p) + (4 \cdot 3p) + (5 \cdot p)$ $E(X) = 0 + 4p + 12p + 5p$ $E(X) = 21p$

Step 5: Calculate the final mean

Since $p = \frac{1}{7}$, we substitute this value into the expression for $E(X)$: $E(X) = 21 \cdot \frac{1}{7}$ $E(X) = 3$

Final Answer: p = 1/7, Mean = 3

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Pedagogical Audit
Bloom's Analysis: This is an REMEMBER question because it requires recalling the basic principles of probability distributions (sum of probabilities equals 1) and the formula for calculating the mean of a discrete random variable.
Knowledge Dimension: CONCEPTUAL
Justification: The question tests the understanding of the concepts of probability distribution and how to calculate the mean from it, rather than just recalling facts or performing a routine procedure.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. It directly relates to the syllabus content on probability distributions and expected value.
Justification: The question is a standard application of the formulas and concepts taught in the probability chapter.

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