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
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To find the probability, we need to determine the number of favorable outcomes (the combinations that satisfy the given condition) and the total number of possible outcomes.
Let's first determine the total number of possible outcomes:
Since there are three nonzero digits with no repeated digits, the total number of possibilities for the first digit is 9 (excluding 0).
After choosing the first digit as 2, we are left with 8 options for the second digit (excluding 0 and 2), and 7 options for the third digit (excluding the digits already chosen).
Therefore, the total number of possible outcomes is 9 * 8 * 7 = 504.
Next, let's determine the number of favorable outcomes:
We need the second digit to be even (divisible by 2) given that the first digit is 2. The even digits remaining are 4, 6, and 8.
So, there are 3 options for the second digit and 7 options for the third digit (same as before, excluding the digits already chosen).
Therefore, the number of favorable outcomes is 3 * 7 = 21.
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = favorable outcomes / total outcomes
Probability = 21 / 504
Probability ≈ 0.0417
Therefore, the probability that the second digit is even, given that the first digit is 2, is approximately 0.0417 or 4.17%.