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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.
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.
We need to find the probability for each possible value of X.
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. $$P(X=0) = \frac{6}{36} = \frac{1}{6}$$
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. $$P(X=1) = \frac{10}{36} = \frac{5}{18}$$
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. $$P(X=2) = \frac{8}{36} = \frac{2}{9}$$
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. $$P(X=3) = \frac{6}{36} = \frac{1}{6}$$
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. $$P(X=4) = \frac{4}{36} = \frac{1}{9}$$
X = 5 when the absolute difference between the numbers is 5. The possible outcomes are (1,6), (6,1). So, there are 2 outcomes. $$P(X=5) = \frac{2}{36} = \frac{1}{18}$$
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