Question

question 10 of 10
in a randomly generated list of numbers from 0 to 8, what is the chance that each number will occur?

a. \\(\frac{1}{8}\\)

b. \\(\frac{1}{7}\\)

c. \\(\frac{1}{6}\\)

d. \\(\frac{1}{9}\\)

Explanation:

Step1: Determine total numbers

Numbers from 0 to 8: 0,1,2,3,4,5,6,7,8. Total count: \( 8 - 0 + 1 = 9 \) numbers.

Step2: Calculate probability

Probability of a specific number occurring is \( \frac{1}{\text{total numbers}} = \frac{1}{9} \)? Wait, no—wait, the options have \( \frac{1}{8} \)? Wait, no, let's recheck. Wait, 0 to 8: how many numbers? 0,1,2,3,4,5,6,7,8: that's 9 numbers? But the options have A: \( \frac{1}{8} \), D: \( \frac{1}{8} \)? Wait, maybe I miscounted. Wait, 0 to 8 inclusive: 8 - 0 + 1 = 9? Wait, no, 0 to 8 is 9 numbers (0 is first, 8 is 9th: 0,1,2,3,4,5,6,7,8: 9 numbers). But the options have \( \frac{1}{8} \) (A and D) and \( \frac{1}{7} \), \( \frac{1}{6} \). Wait, maybe the problem is "numbers from 1 to 8"? Then total numbers 8, probability \( \frac{1}{8} \). Maybe a typo. Assuming numbers from 0 to 7? No. Wait, the options: A is \( \frac{1}{8} \), D is \( \frac{1}{8} \)? Wait, the original problem: "in a randomly generated list of numbers from 0 to 8, what is the chance that each number will occur?" So each number (0-8) has equal chance. Total possible outcomes: 9 numbers? But options don't have \( \frac{1}{9} \). Wait, maybe the problem is "numbers from 1 to 8", so 8 numbers. Then probability is \( \frac{1}{8} \), which is option A (or D, but the options: A is \( \frac{1}{8} \), D is \( \frac{1}{8} \)? Wait, the image: A is \( \frac{1}{8} \), D is \( \frac{1}{8} \)? Maybe a formatting error. Assuming total numbers 8 (e.g., 1-8), then probability is \( \frac{1}{8} \), so option A (or D, but likely A is \( \frac{1}{8} \)).

Answer:

A. \( \frac{1}{8} \)