At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.16 and the probability that the flight will be delayed is 0.05. The probability that it will not rain and the flight will leave on time is 0.81. What is the probability that it is not raining if the flight leaves on time? Round your answer to the nearest thousandth.
Let R be the event that it is raining and D be the event that the flight is delayed.
We are given:
P(R) = 0.16
P(D) = 0.05
P(not R and not D) = 0.81
We know that:
P(not R and not D) = P(not R) * P(not D)
From this, we can find:
P(not R) = 0.81 / P(not D) = 0.81 / (1 - P(D)) = 0.81 / 0.95 = 0.8526
Therefore, the probability that it is not raining if the flight leaves on time is approximately 0.853.
wrong it is 0.02
I apologize for the mistake. Let's correct it.
Let R be the event that it is raining and D be the event that the flight is delayed.
We are given:
P(R) = 0.16
P(D) = 0.05
P(not R and not D) = 0.81
We know that:
P(not R and not D) = P(not R) * P(not D)
Given that P(not R and not D) = 0.81, we can say:
0.81 = P(not R) * (1 - P(D))
0.81 = P(not R) * 0.95
P(not R) = 0.81 / 0.95 = 0.8526
Now, we need to find the probability that it is not raining if the flight leaves on time, denoted as P(not R | not D):
P(not R | not D) = P(not R and not D) / P(not D)
P(not R | not D) = 0.81 / 0.95 = 0.8526
Therefore, the correct probability that it is not raining if the flight leaves on time is approximately 0.853 (rounded to the nearest thousandth). Thank you for pointing out the error.
no it is 0.02
Let's reevaluate the problem to determine the correct answer.
We are given:
P(R) = 0.16
P(D) = 0.05
P(not R and not D) = 0.81
We want to find the probability that it is not raining if the flight leaves on time, denoted as P(not R | not D).
We know that:
P(not R and not D) = P(not R) * P(not D)
Using this information:
0.81 = P(not R) * (1 - 0.05)
0.81 = P(not R) * 0.95
P(not R) = 0.81 / 0.95 ≈ 0.8526
Now, we can find the probability that it is not raining if the flight leaves on time:
P(not R | not D) = P(not R and not D) / P(not D)
P(not R | not D) = 0.81 / 0.95 ≈ 0.8526
Therefore, the probability that it is not raining if the flight leaves on time is approximately 0.853 (rounded to three decimal places). Thank you for your feedback.
ugh
Let's approach the problem again to ensure we arrive at the correct solution.
From the given information:
P(R) = 0.16
P(D) = 0.05
P(not R and not D) = 0.81
We know that P(not R and not D) = P(not R) * P(not D)
Substitute the given values:
0.81 = P(not R) * (1 - 0.05)
0.81 = P(not R) * 0.95
P(not R) = 0.81 / 0.95 ≈ 0.8526
Now we want to find the probability that it is not raining given that the flight leaves on time, denoted as P(not R | not D).
We have:
P(not R | not D) = P(not R and not D) / P(not D)
P(not R | not D) = 0.81 / (1 - 0.05)
P(not R | not D) = 0.81 / 0.95
P(not R | not D) ≈ 0.8526
Therefore, the probability that it is not raining if the flight leaves on time is approximately 0.853 (rounded to three decimal places). Thank you for your patience, and I hope this clarification helps.