To solve the problem and find [tex]\( P(B \mid A) \)[/tex], we need to understand a few key concepts:
- Total number of combinations: Since the digits range from 2 to 9 (inclusive), we have 8 digits to choose from. The digits in a combination cannot repeat, so each combination is formed by choosing 2 different digits in a specific order. Hence, the total number of possible combinations is [tex]\( 8 \times 7 = 56 \)[/tex].
- Event [tex]\( A \)[/tex]: The first digit is less than 6. The digits less than 6 are 2, 3, 4, and 5. Thus, there are 4 possible choices for the first digit. For each of these choices, the second digit can be any of the remaining 7 digits (since digits are not repeated). Therefore, the number of favorable outcomes for Event [tex]\( A \)[/tex] is [tex]\( 4 \times 7 = 28 \)[/tex].
- Event [tex]\( B \)[/tex]: The second digit is less than 6 given that the first digit satisfies Event [tex]\( A \)[/tex]. Both digits in Event [tex]\( A \)[/tex] and Event [tex]\( B \)[/tex] must be distinct.
- Event [tex]\( B \vert A \)[/tex]: Here, we determine how many outcomes satisfy Event [tex]\( B \)[/tex] given that Event [tex]\( A \)[/tex] has already occurred. Since the first digit is fixed and is one of 2, 3, 4, or 5, for the second digit to be less than 6 it can be 2, 3, 4, or 5, excluding the first digit. Therefore, there are 3 remaining choices for the second digit for each of the 4 choices of the first digit. Thus, the number of favorable outcomes for Event [tex]\( B \)[/tex] given Event [tex]\( A \)[/tex] is [tex]\( 4 \times 4 = 16 \)[/tex].
Thus, we can now calculate the probability [tex]\( P(B \mid A) \)[/tex] as follows:
[tex]\[
P(B \mid A) = \frac{\text{Number of favorable outcomes for both A and B}}{\text{Number of favorable outcomes for A}} = \frac{16}{28} = \frac{4}{7}
\][/tex]
So, the correct answer is [tex]\( \boxed{\frac{4}{7}} \)[/tex].
