Six fair dice are tossed independently. Find the probability that the number of 1’s minus the number of 2’s will be 3. 
 	

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 \( (X_1, X_2) \).

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



Six fair dice are tossed independently. Find the probability that the number of 1’s minus the number of 2’s will be 3.

Explore how to calculate the probability that the difference between the number of 1's and 2's in six independent dice rolls is 3. This solution covers concepts like binomial distribution, combinatorics, and the use of probability formulas. Ideal for high school students in Grades 10-12.

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