To solve this problem, let's break it down into simple steps:
1. Understand the problem:
- We have a locker combination consisting of two nonzero digits.
- Nonzero digits range from 1 to 9.
- The first number in the combination is already defined as 3.
- We need to determine the probability that the second number is also 3.
2. Analyze the sample space:
- Since the nonzero digits (1 through 9) are the only possible choices, we have 9 possible choices for each digit. This includes 1, 2, 3, 4, 5, 6, 7, 8, and 9.
3. Determine the desired outcome:
- We want the second digit to be 3.
- Out of the 9 possible digits for the second number, only one of these digits meets our desired outcome (which is 3).
4. Calculate the probability:
- The probability \( P \) of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.
- Here, the number of favorable outcomes is 1 (since only one digit, 3, is favorable).
- The total number of possible outcomes for the second digit is 9 (since it can be any nonzero digit from 1 to 9).
[tex]\[
P(\text{second number is 3}) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}} = \frac{1}{9}
\][/tex]
So, the probability that the second number is 3 is \( \frac{1}{9} \).
Hence, the correct answer is:
[tex]\[
\boxed{\frac{1}{9}}
\][/tex]
