Question

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 \( X \).

Hint:

For probability distributions involving dice, systematically count the outcomes for each value and ensure the total probability sums to 1.

Answer 1

Step 1: Define the random variable \( X \).
Let \( X \) be the absolute difference between the numbers shown on two dice. The possible values of \( X \) are: \[ X = 0, 1, 2, 3, 4, 5. \] Step 2: Count favorable outcomes for each \( X \).
The total number of outcomes when two dice are rolled is \( 6 \times 6 = 36 \). We now count the number of outcomes corresponding to each value of \( X \). - For \( X = 0 \): The numbers on the two dice must be the same: \[ (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6). \] There are 6 outcomes. - For \( X = 1 \): The numbers on the two dice differ by 1. Possible outcomes are: \[ (1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5), (5, 4), (5, 6), (6, 5). \] This gives \( 10 \) outcomes. - For \( X = 2 \): The numbers differ by 2. Possible outcomes are: \[ (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4). \] This gives \( 8 \) outcomes. - For \( X = 3 \): The numbers differ by 3. Possible outcomes are: \[ (1, 4), (4, 1), (2, 5), (5, 2), (3, 6), (6, 3). \] This gives \( 6 \) outcomes. - For \( X = 4 \): The numbers differ by 4. Possible outcomes are: \[ (1, 5), (5, 1), (2, 6), (6, 2). \] This gives \( 4 \) outcomes. - For \( X = 5 \): The numbers differ by 5. Possible outcomes are: \[ (1, 6), (6, 1). \] This gives \( 2 \) outcomes. Step 3: Calculate probabilities.
The probabilities are calculated as: \[ P(X = k) = \frac{\text{Number of favorable outcomes for } X = k}{36}. \] Thus: \[ P(X = 0) = \frac{6}{36}, \quad P(X = 1) = \frac{10}{36}, \quad P(X = 2) = \frac{8}{36}, \quad P(X = 3) = \frac{6}{36}, \quad P(X = 4) = \frac{4}{36},\]\[ \quad P(X = 5) = \frac{2}{36}. \] Step 4: Write the probability distribution.
The probability distribution of \( X \) is: \[ P(X = 0) = \frac{1}{6}, \quad P(X = 1) = \frac{5}{18}, \quad P(X = 2) = \frac{2}{9}, \quad P(X = 3) = \frac{1}{6}, \quad P(X = 4) = \frac{1}{9},\]\[ \quad P(X = 5) = \frac{1}{18}. \] Step 5: Verify the total probability.
Add all probabilities: \[ P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = \frac{6}{36} + \frac{10}{36} + \frac{8}{36} + \frac{6}{36} + \frac{4}{36} + \frac{2}{36} = 1. \] The total probability is 1, so the distribution is valid. Conclusion:
The probability distribution of \( X \) is: \[ \boxed{\begin{array}{c|c} X & P(X)
\hline 0 & \frac{1}{6}
1 & \frac{5}{18}
2 & \frac{2}{9}
3 & \frac{1}{6}
4 & \frac{1}{9}
5 & \frac{1}{18}
\end{array}} \]