At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.18 and the probability that the flight will be delayed is 0.13. The probability that it will not rain and the flight will leave on time is 0.8. What is the probability that the flight would leave on time when it is raining? Round your answer to the nearest thousandth.AnswerAttempt 1 out of 2

Let's denote the following events:

• $R$: It will rain.
• $D$: The flight will be delayed.
• $R^c$: It will not rain.
• $D^c$: The flight will leave on time.

Given:

• $P(R) = 0.18$
• $P(D) = 0.13$
• $P(R^c \cap D^c) = 0.8$

We need to find the probability that the flight will leave on time given that it is raining, $P(D^c | R)$.

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

We are given $P(R^c \cap D^c) = 0.8$. We also know: $P(R) = P(R \cap D) + P(R \cap D^c)$ $P(D) = P(R \cap D) + P(R^c \cap D)$

Since $P(R^c) = 1 - P(R) = 1 - 0.18 = 0.82$, we can write: $P(R^c \cap D) = P(R^c) - P(R^c \cap D^c) = 0.82 - 0.8 = 0.02$

Now, we can find $P(R \cap D)$: $P(D) = P(R \cap D) + P(R^c \cap D)$ $0.13 = P(R \cap D) + 0.02$ $P(R \cap D) = 0.13 - 0.02 = 0.11$

Next, we find $P(R \cap D^c)$: $P(R) = P(R \cap D) + P(R \cap D^c)$ $0.18 = 0.11 + P(R \cap D^c)$ $P(R \cap D^c) = 0.18 - 0.11 = 0.07$

Finally, we can find $P(D^c | R)$: $P(D^c | R) = \frac{P(R \cap D^c)}{P(R)} = \frac{0.07}{0.18} \approx 0.389$

So, the probability that the flight will leave on time when it is raining is approximately $0.389$.