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) [tex]\frac{1}{4}[/tex]
B) [tex]\frac{2}{7}[/tex]
C) [tex]\frac{4}{9}[/tex]
D) [tex]\frac{4}{7}[/tex]



Answer :

To determine the probability that the third digit in a locker combination consisting of three unique nonzero digits (1 through 9) is even, given that the first two digits are even, let's break down the problem step-by-step:

1. Identify nonzero digits and even digits:
- There are a total of 9 nonzero digits: {1, 2, 3, 4, 5, 6, 7, 8, 9}.
- Among these, the even digits are 2, 4, 6, and 8. There are 4 even digits in total.

2. Determine the total number of digits after selecting the first two even digits:
- Since the first two digits are even, we select 2 digits from the 4 available even digits.
- After selecting the first two even digits, there are 7 digits remaining (9 total digits - 2 selected digits).

3. Determine the number of remaining even digits:
- We initially had 4 even digits, and we’ve already used 2 of them for the first two positions.
- Therefore, there are 2 remaining even digits.

4. Calculate the probability that the third digit is even:
- The total number of remaining digits is 7.
- The remaining even digits among these 7 are 2.

The probability that the third digit is even is given by the ratio of the number of remaining even digits to the total number of remaining digits:
[tex]\[ \text{Probability} = \frac{\text{Number of remaining even digits}}{\text{Total number of remaining digits}} = \frac{2}{7} \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{\frac{2}{7}} \][/tex]