We are tasked with finding the probability that the number of 1's minus the number of 2's equals 3 when six fair dice are tossed independently.

Let the random variables $X_1$ and $X_2$ represent the number of 1's and the number of 2's that appear in the tosses, respectively. The problem asks us to find the probability that:

$X_1 - X_2 = 3$

Step 1: Analyze the random variables

• Each die has 6 faces, with each face (1, 2, 3, 4, 5, 6) having a probability of $\frac{1}{6}$ of appearing.
• The probability of getting a 1 on any die is $P(\text{1}) = \frac{1}{6}$.
• The probability of getting a 2 on any die is $P(\text{2}) = \frac{1}{6}$.
• The probability of getting any other number (3, 4, 5, or 6) on a die is $P(\text{not 1 or 2}) = \frac{4}{6} = \frac{2}{3}$.

Thus, the variables $X_1$ and $X_2$ represent binomial distributions because each die has two possible outcomes for the number of 1's and 2's.

Step 2: Set up the system

• $X_1 \sim \text{Binomial}(6, \frac{1}{6})$, i.e., the number of 1's follows a binomial distribution with 6 trials and probability $\frac{1}{6}$.
• $X_2 \sim \text{Binomial}(6, \frac{1}{6})$, i.e., the number of 2's follows the same binomial distribution.

We need to compute the probability that $X_1 - X_2 = 3$.

Step 3: Possible pairs $(X_1, X_2)$

We need to find pairs $(X_1, X_2)$ such that $X_1 - X_2 = 3$. This can be written as:

$X_1 = X_2 + 3$

Since both $X_1$ and $X_2$ are bounded by the number of dice (6 in total), we must have:

$X_2 + 3 \leq 6 \quad \Rightarrow \quad X_2 \leq 3$

Thus, the possible values for $X_2$ are 0, 1, 2, or 3. For each value of $X_2$, the corresponding $X_1$ is $X_2 + 3$.

Step 4: Calculate probabilities for each pair

We now compute the probabilities for each valid pair $(X_1, X_2)$.

• $P(X_1 = 3, X_2 = 0)$
  • $P(X_1 = 3) = \binom{6}{3} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^3$
  • $P(X_2 = 0) = \binom{6}{0} \left(\frac{1}{6}\right)^0 \left(\frac{5}{6}\right)^6$