Q.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 $X$.

Concept: Probability distribution of a discrete random variable — listing all possible values of $X$ and their probabilities.

Step 1 – Possible values of $X$

When two dice are thrown, the absolute difference $X$ can be $0, 1, 2, 3, 4, 5$.

Step 2 – Count favourable outcomes for each $X$

Total outcomes = $6 \times 6 = 36$.

• $X = 0$: same number on both dice → $(1,1), (2,2), \dots, (6,6)$ → 6 outcomes.
• $X = 1$: pairs differing by 1 → $(1,2), (2,1), (2,3), (3,2), \dots, (5,6), (6,5)$ → 10 outcomes.
• $X = 2$: pairs differing by 2 → 8 outcomes.
• $X = 3$: pairs differing by 3 → 6 outcomes.
• $X = 4$: pairs differing by 4 → 4 outcomes.
• $X = 5$: pairs differing by 5 → $(1,6), (6,1)$ → 2 outcomes.

Step 3 – Probability for each $X$

Divide each count by 36.

| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |

|-----|---|---|---|---|---|---| …

Concept and Intuition

When two dice are thrown, each die shows a number from 1 to 6. The total number of ordered pairs $(a,b)$ is $6 \times 6 = 36$, all equally likely. The absolute difference $X = |a-b|$ can range from 0 (when both dice show the same number) to 5 (when one die shows 1 and the other shows 6).

The key insight: instead of listing all 36 pairs blindly, we can systematically count how many pairs give each difference. For a given difference $d$, the pairs $(a,b)$ satisfy $|a-b| = d$. This means $a-b = d$ or $a-b = -d$, i.e., $b = a-d$ or $b = a+d$, with both $a$ and $b$ between 1 and 6.

Step-by-Step Solution

1. Understand the sample space.

Each die is independent. The sample space has $6 \times 6 = 36$ ordered outcomes: $(1,1), (1,2), \dots, (6,6)$. Every outcome has probability $\frac{1}{36}$.

2. Define the random variable.

$X = |a-b|$, where $a$ is the number on the first die and $b$ on the second. $X$ can take values $0, 1, 2, 3, 4, 5$.

3. Count outcomes for $X=0$.

$|a-b|=0$ means $a=b$. The pairs are $(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)$. That's 6 outcomes.

So $P(X=0) = \frac{6}{36}$.

4. Count outcomes for $X=1$.

$|a-b|=1$ means $a$ and $b$ differ by exactly 1.

• If $a-b=1$: $b = a-1$, so $a$ can be 2 to 6 → pairs: $(2,1), (3,2), (4,3), (5,4), (6,5)$ → 5 outcomes.
• If $a-b=-1$: $b = a+1$, so $a$ can be 1 to 5 → pairs: $(1,2), (2,3), (3,4), (4,5), (5,6)$ → 5 outcomes. Total = $5+5 = 10$. So $P(X=1) = \frac{10}{36}$.

5. Count outcomes for $X=2$.

$|a-b|=2$:

• $a-b=2$: $b = a-2$, $a$ from 3 to 6 → 4 outcomes: $(3,1), (4,2), (5,3), (6,4)$.
• $a-b=-2$: $b = a+2$, $a$ from 1 to 4 → 4 outcomes: $(1,3), (2,4), (3,5), (4,6)$. Total = $4+4 = 8$. So $P(X=2) = \frac{8}{36}$.

6. Count outcomes for $X=3$.

$|a-b|=3$:

• $a-b=3$: $b = a-3$, $a$ from 4 to 6 → 3 outcomes: $(4,1), (5,2), (6,3)$.
• $a-b=-3$: $b = a+3$, $a$ from 1 to 3 → 3 outcomes: $(1,4), (2,5), (3,6)$. Total = $3+3 = 6$. So $P(X=3) = \frac{6}{36}$.

7. Count outcomes for $X=4$.

$|a-b|=4$:

• $a-b=4$: $b = a-4$, $a$ from 5 to 6 → 2 outcomes: $(5,1), (6,2)$. …

Method: Building a Probability Distribution by Counting Outcomes

Use this to construct the distribution of a random variable defined on an equally-likely sample space.

Steps

Step 1: List every value the variable can take.

Determine the smallest and largest possible values of $X$ and everything in between.

Step 2: Count the favourable outcomes for each value.

Work over the full equally-likely sample space. For two dice there are $6\times6=36$ ordered outcomes, each of probability $\tfrac1{36}$. Count how many ordered pairs produce each value of $X$. …

Common Mistakes

Mistake 1: Treating $(a,b)$ and $(b,a)$ as one outcome.

Why it's wrong: the $36$ outcomes are ordered, so a difference of $1$ arises from both $(2,1)$ and $(1,2)$. Merging them halves the count. Correct approach: count ordered pairs — $X=1$ has $10$, not $5$.

Mistake 2: Counting $X=0$ as $12$ (using $2\times6$). …