At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.08 and the probability that the flight will be delayed is 0.09. The probability that it will not rain and the flight will leave on time is 0.87. What is the probability that the flight would be delayed when it is not raining? Round your answer to the nearest thousandth.

Let's denote the following events:

• $R$: It will rain.
• $D$: The flight will be delayed.

Given:

• $P(R) = 0.08$
• $P(D) = 0.09$
• $P(\neg R \cap \neg D) = 0.87$

We need to find the probability that the flight will be delayed given that it is not raining, $P(D \mid \neg R)$.

First, we use the fact that the total probability must sum to 1: $P(R \cap D) + P(R \cap \neg D) + P(\neg R \cap D) + P(\neg R \cap \neg D) = 1$

We know: $P(\neg R \cap \neg D) = 0.87$

Next, we calculate $P(\neg R)$: $P(\neg R) = 1 - P(R) = 1 - 0.08 = 0.92$

We also know: $P(\neg R) = P(\neg R \cap D) + P(\neg R \cap \neg D)$

So: $0.92 = P(\neg R \cap D) + 0.87$

Solving for $P(\neg R \cap D)$: $P(\neg R \cap D) = 0.92 - 0.87 = 0.05$

Now, we need to find $P(D \mid \neg R)$: $P(D \mid \neg R) = \frac{P(\neg R \cap D)}{P(\neg R)}$

Substituting the values: $P(D \mid \neg R) = \frac{0.05}{0.92} \approx 0.054$

Rounding to the nearest thousandth: $P(D \mid \neg R) \approx 0.054$

So, the probability that the flight would be delayed when it is not raining is approximately 0.054.