A locker combination has three nonzero digits, and digits cannot be repeated. If the first digit is odd, what is the probability that the second digit is even?

the answer is not 1/8

To find the probability that the second digit is even given that the first digit is odd, we need to determine the total number of possible combinations and the number of favorable combinations where the second digit is even.

First, consider the choices for the first digit. Since it cannot be zero and must be odd, there are 5 possible choices: 1, 3, 5, 7, and 9.

Next, consider the choices for the second digit. Since the first digit is already chosen and it cannot be repeated, there are 4 remaining digits (0, 2, 4, 6). Out of these, two are even (0 and 2) and two are odd (4 and 6).

So, the total number of possible combinations is 5 choices for the first digit multiplied by 4 choices for the second digit, which gives us a total of 20 possible combinations.

Now, let's determine the favorable combinations where the second digit is even. Since there are two even digits (0 and 2) remaining out of the four choices for the second digit, the number of favorable combinations is 5 choices for the first digit multiplied by 2 choices for the second digit, which gives us 10 favorable combinations.

Therefore, the probability that the second digit is even given that the first digit is odd is 10 favorable combinations out of 20 total possible combinations, which simplifies to 10/20 or 1/2.

So, the probability is 1/2 or 0.5.