To find [tex]\( P(A \text{ and } B) \)[/tex], we need to determine the probability that both the first and the second digits of the locker combination are even, given that the digits are different and non-zero.
First, let's determine [tex]\( P(A) \)[/tex], the probability that the first digit (event [tex]\( A \)[/tex]) is even.
1. Determine the set of non-zero digits:
The non-zero digits are [tex]\( \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \)[/tex].
2. Possible even digits among non-zero digits:
The even digits are [tex]\( \{2, 4, 6, 8\} \)[/tex].
3. Calculate [tex]\( P(A) \)[/tex]:
There are 4 even digits out of the 9 non-zero digits. So, the probability that the first digit is an even digit is:
[tex]\[
P(A) = \frac{\text{Number of even digits}}{\text{Total non-zero digits}} = \frac{4}{9}
\][/tex]
Next, let's determine [tex]\( P(B|A) \)[/tex], the conditional probability that the second digit (event [tex]\( B \)[/tex]) is even, given that the first digit (event [tex]\( A \)[/tex]) is already chosen and is different from the second digit.
4. Determine the remaining digits:
Once the first digit is chosen as an even number, there are 8 digits left to choose from for the second digit.
5. Calculate [tex]\( P(B|A) \)[/tex]:
There are still 3 even digits left out of the remaining 8 digits (since one even digit is already used). So, the probability that the second digit is even is:
[tex]\[
P(B|A) = \frac{\text{Number of remaining even digits}}{\text{Total remaining digits}} = \frac{3}{8}
\][/tex]
Using the multiplication rule of conditional probabilities, we can find [tex]\( P(A \text{ and } B) \)[/tex]:
[tex]\[
P(A \text{ and } B) = P(A) \times P(B|A)
\][/tex]
Substituting the values we calculated:
[tex]\[
P(A \text{ and } B) = \left(\frac{4}{9}\right) \times \left(\frac{3}{8}\right) = \frac{1}{6}
\][/tex]
Therefore, the probability that both digits are even is [tex]\( \frac{1}{6} \)[/tex].
The correct answer is:
A) [tex]\(\frac{1}{6}\)[/tex]
