A locker combination has three nonzero digits, and digits cannot be repeated. If the first two digits are even, what is the probability that the third digit is even?
A) 7/9
B) 1/3
C) 1/6
D) 1/7
there are only 4 even nonzero digits
2 are already used, leaving 2 choices possible
there are 5 odd digits, making 7 choices in all.
??
To find the probability that the third digit is even given that the first two digits are even, we first need to determine the total number of possible combinations meeting the given criteria.
We know that the first two digits are even, meaning they can only be 2, 4, 6, or 8.
To find the number of possible options for the first digit, we have 4 choices (2, 4, 6, or 8). Once we choose the first digit, we have three choices left for the second digit (since we cannot repeat digits), leaving us with 4 * 3 = 12 possible combinations for the first two digits.
Now, we need to determine how many options we have for the third digit, given that it must be even. With two even digits already used, we are left with 0, 2, 4, 6, or 8 as options. However, we cannot repeat any digits, so we have 4 choices remaining for the third digit.
Therefore, the total number of possible combinations meeting the given criteria is 12 * 4 = 48.
Now, let's calculate the probability that the third digit is even. We are interested in the number of combinations where the third digit is even, divided by the total number of possible combinations.
Out of the 48 possible combinations, 24 will have an even third digit (since half of the numbers from 0 to 9 are even).
Therefore, the probability that the third digit is even given that the first two digits are even is 24/48 = 1/2.
However, we need to choose from the given answer options:
A) 7/9
B) 1/3
C) 1/6
D) 1/7
None of the given options match the calculated probability of 1/2. Therefore, the correct answer is not provided in the given options.