To determine the correct answer to this question, let's analyze the possible outcomes and their associated probabilities.
In a randomly generated list of numbers from 0 to 5, the numbers include 0, 1, 2, 3, 4, and 5. This gives us a total of 6 different numbers.
For each number to occur, we must consider the total number of equally likely outcomes. Here, the numbers range from 0 to 5, giving us six possible outcomes.
The probability of each number occurring in this list is calculated by taking the ratio of the number of favorable outcomes to the total number of possible outcomes. Given that each number can occur with equal probability:
[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]
Since there are 6 different numbers and each number can occur only once:
[tex]\[ \text{Probability of each number occurring} = \frac{1}{6} \][/tex]
We are given a claim that the chance that each number can occur is [tex]\(\frac{1}{5}\)[/tex]. However, we have calculated that the correct probability for each number from 0 to 5 occurring is actually [tex]\(\frac{1}{6}\)[/tex].
Thus, the correct answer to the question is:
B. False
