Let \(E\), \(L\), and \(OT\) represent the events that Ana is early, late, and on time, respectively. Let \(B\) represent the event that Ana eats a banana, and \(\overline{B}\) represent the event that she does not eat a banana.

We need to find \(P(OT | \overline{B})\).

Using the law of total probability, we have:

\(P(\overline{B}) = P(\overline{B} | E)P(E) + P(\overline{B} | L)P(L) + P(\overline{B} | OT)P(OT)\)

Given:

• \(P(E) = 0.05\)
• \(P(L) = 0.75\)
• \(P(OT) = 1 - P(E) - P(L) = 0.2\)
• \(P(\overline{B} | E) = 1 - 0.7 = 0.3\)
• \(P(\overline{B} | L) = 1\)
• \(P(\overline{B} | OT) = 1 - 0.4 = 0.6\)

Substitute these values into the equation:

\(P(\overline{B}) = 0.3 \times 0.05 + 1 \times 0.75 + 0.6 \times 0.2\)

\(P(\overline{B}) = 0.015 + 0.75 + 0.12 = 0.885\)

Now, use Bayes' theorem to find \(P(OT | \overline{B})\):

\(P(OT | \overline{B}) = \frac{P(\overline{B} | OT)P(OT)}{P(\overline{B})} = \frac{0.6 \times 0.2}{0.885}\)

\(P(OT | \overline{B}) = \frac{0.12}{0.885} \approx 0.136\)

Thus, the probability that Ana is on time given that she does not eat a banana is \(0.136 \left( \frac{8}{59} \right)\).