To determine the probability that the difference between the highest and lowest numbers on three fair dice is exactly four, we need to consider all the possible outcomes and how many of these outcomes meet our condition.

Firstly, we know that the highest and lowest numbers on the dice must differ by exactly four. Since each die has faces numbered from 1 to 6, the possible pairs for the highest and lowest numbers are:

• (5, 1)
• (6, 2)

These pairs satisfy the condition of having a difference of four. We now need to count the number of ways to achieve each pair using three dice.

Pair (5, 1)

For the dice to show a highest number of 5 and a lowest number of 1, the possible numbers on the dice can be 1, 5, and any number between 1 and 5 inclusive. Therefore, the third die must show any number from 1 to 5, which are 1, 2, 3, 4, 5.

We need to count all permutations of these outcomes:

• Dice showing (1, 1, 5)
• Dice showing (1, 2, 5)
• Dice showing (1, 3, 5)
• Dice showing (1, 4, 5)
• Dice showing (1, 5, 5)

The number of permutations for each scenario is as follows:

1. (1, 1, 5): $\frac{3!}{2!1!} = 3$ ways
2. (1, 2, 5): $\frac{3!}{1!1!1!} = 6$ ways
3. (1, 3, 5): $\frac{3!}{1!1!1!} = 6$ ways
4. (1, 4, 5): $\frac{3!}{1!1!1!} = 6$ ways
5. (1, 5, 5): $\frac{3!}{1!2!} = 3$ ways

So, the total number of permutations for the pair (5, 1) is: $3 + 6 + 6 + 6 + 3 = 24$

Pair (6, 2)

For the dice to show a highest number of 6 and a lowest number of 2, the possible numbers on the dice can be 2, 6, and any number between 2 and 6 inclusive. Therefore, the third die must show any number from 2 to 6, which are 2, 3, 4, 5, 6.

We need to count all permutations of these outcomes:

• Dice showing (2, 2, 6)
• Dice showing (2, 3, 6)
• Dice showing (2, 4, 6)
• Dice showing (2, 5, 6)
• Dice showing (2, 6, 6)

The number of permutations for each scenario is as follows:

1. (2, 2, 6): $\frac{3!}{2!1!} = 3$ ways
2. (2, 3, 6): $\frac{3!}{1!1!1!} = 6$ ways
3. (2, 4, 6): $\frac{3!}{1!1!1!} = 6$ ways
4. (2, 5, 6): $\frac{3!}{1!1!1!} = 6$ ways
5. (2, 6, 6): $\frac{3!}{1!2!} = 3$ ways

So, the total number of permutations for the pair (6, 2) is: $3 + 6 + 6 + 6 + 3 = 24$

Total Favorable Outcomes

The total number of favorable outcomes is the sum of the outcomes for both pairs: $24 + 24 = 48$

Total Possible Outcomes

The total number of outcomes when rolling three dice is: $6 \times 6 \times 6 = 216$

Probability Calculation

The probability is the ratio of the number of favorable outcomes to the total number of possible outcomes: $P = \frac{48}{216} = \frac{2}{9} \approx 0.222$

Thus, the probability that the difference between the highest and lowest numbers showing on three dice is exactly four is: $\boxed{0.222}$

Would you like more details or have any questions?

Here are 5 more questions you can ask:

1. How do you calculate the probability of rolling a specific sum with two dice?
2. What is the probability of getting at least one six when rolling four dice?
3. How do you find the expected value of a random variable?
4. Can you explain the concept of conditional probability with an example?
5. How do you solve a system of linear equations using matrix methods?

Tip: When solving probability problems, always consider breaking down the problem into smaller parts and analyzing each part step-by-step for a clear solution.