Question

Two dice are thrown simultaneously. Let $X$ be the random variable representing the absolute difference of the numbers appeared on the dice. The mean of $X$ is:

• A. $\frac{35}{18}$
• B. $\frac{13}{36}$
• C. $\frac{7}{12}$
• D. $3$

Hint:

For absolute differences of two dice, the number of outcomes always follows a clear linear progression: there are $6$ outcomes for a difference of $0$, and $2(6 - d)$ outcomes for any difference $d > 0$.

Answer 1

The Correct Option is A

Concept: When two fair dice are rolled, the total number of outcomes in the sample space is $6 \times 6 = 36$. Each outcome can be written as an ordered pair $(i, j)$, where $1 \le i, j \le 6$. The random variable $X$ is defined as the absolute difference: $X = |i - j|$. The possible values that $X$ can take are $0, 1, 2, 3, 4, 5$. The mean (expected value) of $X$ is computed using the formula: \[ \mu = E[X] = \sum x_i \cdot P(X = x_i) \]

Step 1: Find the frequency and probability distribution for each value of $X$.
Let us count the pairs corresponding to each difference: * For $X = 0$: Pairs where $i = j$. These are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6) \implies 6$ pairs. \[ P(X = 0) = \frac{6}{36} \] * For $X = 1$: Pairs where $|i - j| = 1$. These are $(1,2), (2,3), (3,4), (4,5), (5,6)$ and their reverses $\implies 5 \times 2 = 10$ pairs. \[ P(X = 1) = \frac{10}{36} \] * For $X = 2$: Pairs where $|i - j| = 2$. These are $(1,3), (2,4), (3,5), (4,6)$ and their reverses $\implies 4 \times 2 = 8$ pairs. \[ P(X = 2) = \frac{8}{36} \] * For $X = 3$: Pairs where $|i - j| = 3$. These are $(1,4), (2,5), (3,6)$ and their reverses $\implies 3 \times 2 = 6$ pairs. \[ P(X = 3) = \frac{6}{36} \] * For $X = 4$: Pairs where $|i - j| = 4$. These are $(1,5), (2,6)$ and their reverses $\implies 2 \times 2 = 4$ pairs. \[ P(X = 4) = \frac{4}{36} \] * For $X = 5$: Pairs where $|i - j| = 5$. These are $(1,6)$ and its reverse $(6,1) \implies 1 \times 2 = 2$ pairs. \[ P(X = 5) = \frac{2}{36} \] Let us double check the sum of pairs: $6 + 10 + 8 + 6 + 4 + 2 = 36$. The count is correct.

Step 2: Calculate the expected value $E[X]$.
\[ E[X] = (0)\left(\frac{6}{36}\right) + (1)\left(\frac{10}{36}\right) + (2)\left(\frac{8}{36}\right) + (3)\left(\frac{6}{36}\right) + (4)\left(\frac{4}{36}\right) + (5)\left(\frac{2}{36}\right) \] \[ = \frac{0 + 10 + 16 + 18 + 16 + 10}{36} \] \[ = \frac{70}{36} = \frac{35}{18} \]