Class CBSE Class 12 Mathematics Probability Q #1280
KNOWLEDGE BASED
UNDERSTAND
3 Marks 2024 AISSCE(Board Exam) SA
A pair of dice is thrown simultaneously. If X denotes the absolute difference of the numbers appearing on top of the dice, then find the probability distribution of X.
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Detailed Solution

Step 1: Define the Sample Space

When a pair of dice is thrown, the sample space consists of 36 possible outcomes. Let S be the sample space. Then, n(S) = 36.

Step 2: Define the Random Variable X

X denotes the absolute difference of the numbers appearing on the top of the dice. The possible values of X are 0, 1, 2, 3, 4, and 5.

Step 3: Calculate the Probabilities for each value of X

We need to find the probability for each possible value of X.

Step 4: Calculate P(X=0)

X = 0 when the numbers on both dice are the same. The possible outcomes are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, there are 6 outcomes. $$P(X=0) = \frac{6}{36} = \frac{1}{6}$$

Step 5: Calculate P(X=1)

X = 1 when the absolute difference between the numbers is 1. The possible outcomes 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 outcomes. $$P(X=1) = \frac{10}{36} = \frac{5}{18}$$

Step 6: Calculate P(X=2)

X = 2 when the absolute difference between the numbers is 2. The possible outcomes are (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4). So, there are 8 outcomes. $$P(X=2) = \frac{8}{36} = \frac{2}{9}$$

Step 7: Calculate P(X=3)

X = 3 when the absolute difference between the numbers is 3. The possible outcomes are (1,4), (4,1), (2,5), (5,2), (3,6), (6,3). So, there are 6 outcomes. $$P(X=3) = \frac{6}{36} = \frac{1}{6}$$

Step 8: Calculate P(X=4)

X = 4 when the absolute difference between the numbers is 4. The possible outcomes are (1,5), (5,1), (2,6), (6,2). So, there are 4 outcomes. $$P(X=4) = \frac{4}{36} = \frac{1}{9}$$

Step 9: Calculate P(X=5)

X = 5 when the absolute difference between the numbers is 5. The possible outcomes are (1,6), (6,1). So, there are 2 outcomes. $$P(X=5) = \frac{2}{36} = \frac{1}{18}$$

Step 10: Summarize the Probability Distribution

The probability distribution of X is: X = 0: P(X=0) = 1/6 X = 1: P(X=1) = 5/18 X = 2: P(X=2) = 2/9 X = 3: P(X=3) = 1/6 X = 4: P(X=4) = 1/9 X = 5: P(X=5) = 1/18

Final Answer: X = 0: P(X=0) = 1/6, X = 1: P(X=1) = 5/18, X = 2: P(X=2) = 2/9, X = 3: P(X=3) = 1/6, X = 4: P(X=4) = 1/9, X = 5: P(X=5) = 1/18

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because the student needs to understand the concept of probability distribution and apply it to the given scenario of throwing a pair of dice. They need to comprehend how to calculate the absolute difference and then determine the probabilities for each possible value.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of the concepts of random variables, sample space, and probability distribution, rather than just recalling facts or following a specific algorithm.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly relates to the syllabus content on probability distributions and random variables, typically found in the Probability chapter. It tests the student's understanding of these concepts through a standard textbook problem.

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