To solve this problem, let's break it down step-by-step:
1. Determine the total number of nonzero digits:
- The digits available are 1, 2, 3, 4, 5, 6, 7, 8, and 9, totaling 9 digits.
2. Identify the odd digits:
- The odd digits among these are 1, 3, 5, 7, and 9, giving us a total of 5 odd digits.
3. Understand the condition given:
- The first two digits of the locker combination are odd, and they cannot be repeated. Hence, two out of the five odd digits have already been used.
4. Calculate the remaining odd digits after the first two are chosen:
- Since two odd digits are used up, we are left with 3 odd digits.
5. Determine the remaining total digits available:
- Initially, there were 9 digits. Since the first two have been used, there are 7 digits left.
6. Find the probability that the third digit is also odd:
- The number of favorable outcomes (choosing an odd digit) is 3.
- The number of possible outcomes (choosing any remaining digit) is 7.
- Therefore, the probability that the third digit is odd is the ratio of remaining odd digits to the total remaining digits:
[tex]\[
\text{Probability} = \frac{\text{Number of remaining odd digits}}{\text{Total remaining digits}} = \frac{3}{7}
\][/tex]
So, the probability that the third digit is also odd, given that the first two digits are odd, is \(\frac{3}{7}\).
The correct answer is C) [tex]\(\frac{3}{7}\)[/tex].
