To solve this problem, let's break it down step by step.
1. Determine the available digits:
- The first digit has already been chosen as 2. Since the digits must be nonzero and without repetition, we need to exclude 2 from our list of available digits.
- The remaining possible digits for the second and third positions are therefore: 1, 3, 4, 5, 6, 7, 8, and 9.
2. Identify the even digits available for the second position:
- Excluding 2, the remaining even digits in the list of available digits are 4, 6, and 8.
3. Calculate the probability:
- The total number of available digits for the second position is 8 (since excluding 2 from the original 9 digits leaves us with 8 digits).
- Out of these 8 possible digits, 3 are even (4, 6, and 8).
- Therefore, the probability that the second digit is even is the number of even digits divided by the total number of possible digits for the second position.
So, the probability is calculated as:
[tex]\[
\text{Probability} = \frac{\text{Number of even digits}}{\text{Total possible digits}} = \frac{3}{8}
\][/tex]
Thus, the probability that the second digit is also even is [tex]\(\boxed{\frac{3}{8}}\)[/tex].
