A locker combination has three nonzero digits, with no repeated digits. If the first digit is a 2, what is the probability the second digit is also even?

3/8

1/4

To find the probability that the second digit of a locker combination is even, given that the first digit is 2, we first need to determine the total number of possible combinations when the first digit is 2.

Since there are three nonzero digits, two of them must be chosen from the remaining eight digits (0, 1, 3, 4, 5, 6, 7, 8, 9), and they cannot be repeated. The number of ways to choose two digits out of eight is calculated using the combination formula, which is nCr = n! / (r!(n-r)!), where n is the total number of digits and r is the number of digits to be chosen.

So, in our case, there are 8 digits to choose from and we need to choose 2, so the number of ways to choose the remaining two digits is 8C2 = 8! / (2!(8-2)!) = 8! / (2!6!) = (8 * 7) / (2 * 1) = 28.

Now that we know there are 28 possible combinations for the remaining two digits when the first digit is 2, we can determine the number of combinations where the second digit is even. Since there are five even digits (0, 2, 4, 6, 8), there are four even digits left to choose from for the second digit (excluding 2 itself).

Therefore, the number of combinations where the second digit is even is 4, and the probability can be calculated by dividing this number by the total number of possible combinations:

Probability = Number of favorable outcomes / Total number of outcomes
Probability = 4 / 28
Probability = 1 / 7

So, the probability that the second digit of the locker combination is even, given that the first digit is 2, is 1/7.