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consider the experiment of rolling two dice let x denotes the absolute difference of outcomes find the pmf of X

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Solution

The experiment involves rolling two dice. Let's denote X as the absolute difference of the outcomes. The possible values of X can range from 0 (when both dice show the same number) to 5 (when one die shows 6 and the other shows 1).

Step 1: Identify the sample space The sample space S of two dice is the set of all possible outcomes. Since a die has 6 faces, there are 6*6 = 36 possible outcomes when two dice are rolled.

Step 2: Identify the possible values of X The possible values of X are {0, 1, 2, 3, 4, 5}.

Step 3: Calculate the probability mass function (pmf) The pmf of a discrete random variable is the function that gives the probability that a random variable is exactly equal to some value.

Let's calculate the pmf for each possible value of X:

  • P(X=0): This occurs when both dice show the same number. There are 6 such outcomes (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). So, P(X=0) = 6/36 = 1/6.

  • P(X=1): This occurs when the numbers on the dice differ by 1. There are 10 such outcomes (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,5), (5,4), (5,6), (6,5). So, P(X=1) = 10/36 = 5/18.

  • P(X=2): This occurs when the numbers on the dice differ by 2. There are 8 such outcomes (1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4). So, P(X=2) = 8/36 = 2/9.

  • P(X=3): This occurs when the numbers on the dice differ by 3. There are 6 such outcomes (1,4), (4,1), (2,5), (5,2), (3,6), (6,3). So, P(X=3) = 6/36 = 1/6.

  • P(X=4): This occurs when the numbers on the dice differ by 4. There are 4 such outcomes (1,5), (5,1), (2,6), (6,2). So, P(X=4) = 4/36 = 1/9.

  • P(X=5): This occurs when the numbers on the dice differ by 5. There are 2 such outcomes (1,6), (6,1). So, P(X=5) = 2/36 = 1/18.

So, the pmf of X is: P(X=x) = {1/6, 5/18, 2/9, 1/6, 1/9, 1/18} for x = {0, 1, 2, 3, 4, 5}.

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