Let the event that the first digit is 5 be denoted as \(5_1\) and the event that the second digit is not 5 be denoted as \(\overline{5}_2\).

The probability that the first digit is 5 and the second digit is not 5 is:

\(P(5_1 \cap \overline{5}_2) = \frac{3}{9} \times \frac{6}{8} = \frac{1}{4}\)

The probability that the second digit is not 5 is:

\(P(\overline{5}_2) = \frac{1}{4} + \frac{6}{9} \times \frac{5}{8} = \frac{48}{72} = 0.6666\)

The conditional probability that the first digit is 5 given that the second digit is not 5 is:

\(P(5_1 | \overline{5}_2) = \frac{\frac{1}{4}}{\frac{48}{72}} = \frac{3}{8} = 0.375\)