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 0.14. the probability that it will not rain and the flight will leave on time is 0.76. what is the probability that the flight would be delayed when it is raining? round your answer to the nearest thousandth.

Explanation:

Step1: Define events

Let \( R = \text{rain} \), \( D = \text{flight delayed} \). Given \( P(R)=0.11 \), \( P(D)=0.14 \), \( P(
eg R \cap
eg D)=0.76 \).

Step2: Find \( P(

eg R) \)
\( P(
eg R) = 1 - P(R) = 1 - 0.11 = 0.89 \).

Step3: Find \( P(

eg D|
eg R) \)
\( P(
eg D|
eg R) = \frac{P(
eg R \cap
eg D)}{P(
eg R)} = \frac{0.76}{0.89} \approx 0.8539 \).

Step4: Find \( P(D|

eg R) \)
\( P(D|
eg R) = 1 - P(
eg D|
eg R) \approx 1 - 0.8539 = 0.1461 \).

Step5: Use total probability for \( P(D) \)

\( P(D) = P(D|R)P(R) + P(D|
eg R)P(
eg R) \). Substitute values:
\( 0.14 = P(D|R)(0.11) + 0.1461(0.89) \).
Calculate \( 0.1461×0.89 ≈ 0.1300 \).
Then \( 0.14 = 0.11P(D|R) + 0.1300 \).
Solve for \( P(D|R) \):
\( 0.11P(D|R) = 0.14 - 0.1300 = 0.01 \)? No correction: Wait, correct total probability:
Wait \( P(D) = P(D∩R) + P(D∩¬R) \), and \( P(¬R∩¬D)=0.76 \), so \( P(¬R) = P(¬R∩D) + P(¬R∩¬D) \) → \( P(¬R∩D)=0.89 - 0.76=0.13 \). Then \( P(D)=P(D∩R)+P(D∩¬R) \) → \( 0.14=P(D∩R)+0.13 \) → \( P(D∩R)=0.01 \)? No wait no: \( P(D∩¬R)=P(¬R)-P(¬R∩¬D)=0.89-0.76=0.13 \). Then \( P(D)=P(D∩R)+0.13=0.14 \) → \( P(D∩R)=0.01 \)? No mistake earlier: Correct step: \( P(D∩¬R)=P(¬R)-P(¬R∩¬D)=0.89-0.76=0.13 \). Then \( P(D)=P(D∩R)+P(D∩¬R) \) → \( 0.14 = P(D∩R) +0.13 \) → \( P(D∩R)=0.01 \)? No wait no: Wait \( P(D)=0.14 \), so \( P(D∩R)=P(D)-P(D∩¬R)=0.14-0.13=0.01 \). Then \( P(D|R)=P(D∩R)/P(R)=0.01/0.11≈0.0909 \)? No correction: Wait no, wrong: \( P(¬R∩¬D)=0.76 \), so \( P(R∪D)=1-P(¬R∩¬D)=0.24 \). And \( P(R∪D)=P(R)+P(D)-P(R∩D) \). So \( 0.24=0.11+0.14-P(R∩D) \) → \( P(R∩D)=0.11+0.14-0.24=0.01 \). Then \( P(D|R)=0.01/0.11≈0.091 \)? No wait original problem: Wait maybe I misread: The problem says "probability that it will not rain and the flight will leave on time is 0.76". So \( P(¬R ∩ ¬D)=0.76 \). Then \( P(R)=0.11 \), so \( P(¬R)=0.89 \). \( P(¬R)=P(¬R∩D)+P(¬R∩¬D) \) → \( P(¬R∩D)=0.89-0.76=0.13 \). \( P(D)=P(R∩D)+P(¬R∩D)=0.14 \) → \( P(R∩D)=0.14-0.13=0.01 \). Then \( P(D|R)=P(R∩D)/P(R)=0.01/0.11≈0.0909≈0.091 \). But wait why earlier mistake: Yes that's correct. So answer is 0.091? Wait no wait let's check again: \( P(R∪D)=1-P(¬R∩¬D)=0.24 \). \( P(R∪D)=P(R)+P(D)-P(R∩D) \) → \( 0.24=0.11+0.14-P(R∩D) \) → \( P(R∩D)=0.25-0.24=0.01 \). So yes \( P(D|R)=0.01/0.11≈0.0909≈0.091 \).

Wait I think I messed up earlier steps but corrected: The correct answer is approximately 0.091.

Wait no wait another way: Let's do it again carefully:

Given:

• \( P(R) = 0.11 \) (rain)
• \( P(D) = 0.14 \) (delayed)
• \( P(\text{no rain and on time}) = P(

eg R \cap
eg D) = 0.76 \)

We need \( P(D|R) = \frac{P(D \cap R)}{P(R)} \). So we need \( P(D \cap R) \).

First, \( P(
eg R) = 1 - P(R) = 0.89 \).

\( P(
eg R) = P(
eg R \cap D) + P(
eg R \cap
eg D) \) → \( 0.89 = P(
eg R \cap D) + 0.76 \) → \( P(
eg R \cap D) = 0.89 - 0.76 = 0.13 \).

Also, \( P(D) = P(D \cap R) + P(D \cap
eg R) \) → \( 0.14 = P(D \cap R) + 0.13 \) → \( P(D \cap R) = 0.14 - 0.13 = 0.01 \).

Then \( P(D|R) = \frac{0.01}{0.11} ≈ 0.0909 ≈ 0.091 \).

Yes, so the correct answer is 0.091.

Final Answer:

0.091

Step1: Define events

Let \( R=\text{rain}, D=\text{delayed} \). Given values: \( P(R)=0.11, P(D)=0.14, P(
eg R∩¬D)=0.76 \).

Step2: Calculate \( P(

eg R) \)
\( P(
eg R)=1-P(R)=0.89 \).

Step3: Find \( P(

eg R∩D) \)
\( P(
eg R∩D)=P(
eg R)-P(
eg R∩¬D)=0.89-0.76=0.13 \).

Step4: Find \( P(R∩D) \)

\( P(R∩D)=P(D)-P(
eg R∩D)=0.14-0.13=0.01 \).

Step5: Compute \( P(D|R) \)

\( P(D|R)=\frac{P(R∩D)}{P(R)}=\frac{0.01}{0.11}≈0.091 \).

(Note: Earlier confusion was due to miscalculation, corrected steps show the right result.)

Final answer is 0.091.

Answer:

0.636