The problem states:

"4 identical dice are rolled once. The probability that at least 3 different numbers appear on them."

Approach to the Solution:

We need to calculate the probability that at least 3 different numbers appear when rolling 4 dice.

Let's break this down into steps:

1. Total number of possible outcomes: Each die has 6 possible outcomes (numbers from 1 to 6). Since we roll 4 dice, the total number of possible outcomes is: $6^4 = 1296$

2. Number of favorable outcomes (at least 3 different numbers): To calculate this, we will use complementary counting. That is, we'll subtract the cases where fewer than 3 different numbers appear from the total possible outcomes.

   • Case 1: All dice show the same number. There are 6 outcomes where all 4 dice show the same number (one for each number from 1 to 6).

   • Case 2: Exactly two different numbers appear. We need to select 2 numbers from 6, which can be done in $\binom{6}{2} = 15$ ways. Once 2 numbers are chosen, we assign 4 dice to these two numbers. There are $\binom{4}{2} + \binom{4}{3}$ ways to assign two of the dice to one number and the rest to the other. Each of these ways has different multiplicity.

   Finally, subtract these unfavorable outcomes from the total.

Let's calculate the probability.The probability that at least 3 different numbers appear when rolling 4 identical dice is approximately 0.88, or 87.96%.

Would you like more details on any of the steps?

Here are 5 related questions that might interest you:

1. How would the probability change if 5 dice were rolled instead of 4?
2. What is the probability of getting exactly 3 different numbers when rolling 4 dice?
3. How can we calculate the probability for rolling all different numbers on 4 dice?
4. Can we use a similar method for non-identical dice?
5. How would the solution change if the dice had more than 6 faces?

Tip: Complementary counting is a powerful method in probability when it's easier to calculate the complement of the desired outcome.