Class CBSE Class 12 Mathematics Probability Q #911
COMPETENCY BASED
APPLY
3 Marks 2023 SA
A pair of dice is thrown simultaneously. If $X$ denotes the absolute difference of numbers obtained on the pair of dice, then find the probability distribution of $X$.
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Step-by-Step Solution

Let $S$ be the sample space when a pair of dice is thrown. Then $n(S) = 6 \times 6 = 36$.

Let $X$ be the random variable denoting the absolute difference of numbers obtained on the pair of dice. The possible values of $X$ are $0, 1, 2, 3, 4, 5$.

We need to find the probability distribution of $X$, i.e., $P(X=x)$ for $x = 0, 1, 2, 3, 4, 5$.

  • $P(X=0)$: The pairs with absolute difference 0 are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, there are 6 such pairs. $P(X=0) = \frac{6}{36} = \frac{1}{6}$
  • $P(X=1)$: The pairs with absolute difference 1 are (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5). So, there are 10 such pairs. $P(X=1) = \frac{10}{36} = \frac{5}{18}$
  • $P(X=2)$: The pairs with absolute difference 2 are (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4). So, there are 8 such pairs. $P(X=2) = \frac{8}{36} = \frac{2}{9}$
  • $P(X=3)$: The pairs with absolute difference 3 are (1,4), (4,1), (2,5), (5,2), (3,6), (6,3). So, there are 6 such pairs. $P(X=3) = \frac{6}{36} = \frac{1}{6}$
  • $P(X=4)$: The pairs with absolute difference 4 are (1,5), (5,1), (2,6), (6,2). So, there are 4 such pairs. $P(X=4) = \frac{4}{36} = \frac{1}{9}$
  • $P(X=5)$: The pairs with absolute difference 5 are (1,6), (6,1). So, there are 2 such pairs. $P(X=5) = \frac{2}{36} = \frac{1}{18}$

The probability distribution of $X$ is:

$X$ 0 1 2 3 4 5
$P(X)$ $\frac{1}{6}$ $\frac{5}{18}$ $\frac{2}{9}$ $\frac{1}{6}$ $\frac{1}{9}$ $\frac{1}{18}$

Correct Answer: See explanation for probability distribution table.<\/strong>

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply their understanding of probability and absolute difference to construct the probability distribution.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a series of steps to calculate the probabilities for each possible value of the random variable X. This involves a specific algorithm or method.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question requires the student to apply the concepts of probability to a specific scenario, demonstrating their ability to use the knowledge in a practical context rather than just recalling definitions or formulas.

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