Question

find probability given probabilities
score: 1/3 penalty: 1 off
question
at laguardia airport for a certain nightly flight, the probability that it will rain is 0.11 and the probability that the flight will be delayed is 0.15. the probability that it will not rain and the flight will leave on time is 0.81. what is the probability that it is raining if the flight has been delayed? round your answer to the nearest thousandth.
answer attempt 1 out of 2
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Explanation:

Step1: Identify Events and Probabilities

Let \( R \) be the event that it rains, and \( D \) be the event that the flight is delayed. We know:

• \( P(D) = 0.15 \) (probability of flight delay)
• \( P(R) = 0.11 \) (probability of rain)
• \( P(

eg R \cap
eg D) = 0.81 \) (probability of no rain and on - time flight)

First, find \( P(
eg D) \) (probability of flight not delayed). Since \( P(D)+P(
eg D)=1 \), then \( P(
eg D)=1 - P(D)=1 - 0.15 = 0.85 \)

Also, \( P(
eg R \cap
eg D)=0.81 \), and we know that \( P(
eg R \cap
eg D)=P(
eg R)\times P(
eg D|
eg R) \), but we can also find \( P(
eg R) \) from the law of total probability for the non - delayed case.

We know that \( P(
eg D)=P(R\cap
eg D)+P(
eg R\cap
eg D) \). Let \( P(R\cap
eg D)=x \), then \( 0.85=x + 0.81 \), so \( x=P(R\cap
eg D)=0.85 - 0.81 = 0.04 \)

We also know that \( P(R)=P(R\cap D)+P(R\cap
eg D) \). Let \( P(R\cap D)=y \), then \( 0.11=y + 0.04 \), so \( y = P(R\cap D)=0.11 - 0.04 = 0.07 \)

Step2: Apply Bayes' Theorem

We want to find \( P(R|D) \) (probability of rain given flight is delayed). By Bayes' theorem, \( P(R|D)=\frac{P(R\cap D)}{P(D)} \)

We found that \( P(R\cap D) = 0.07 \) and \( P(D)=0.15 \)

So \( P(R|D)=\frac{0.07}{0.15}\approx0.467 \)

Answer:

\( 0.467 \)