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At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.19 and the probability that the flight will be delayed is 0.17 . The probability that it will rain and the flight will be delayed is 0.05 . What is the probability that it is not raining if the flight leaves on time? Round your answer to the nearest thousandth.
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At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 00.1919 and the probability that the flight will be delayed is 00.1717 . The probability that it will rain and the flight will be delayed is 00.0505 . What is the probability that it is not raining if the flight leaves on time? Round your answer to the nearest thousandth.\newlineAnswer:\newlineSubmit Answer

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Q. At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 00.1919 and the probability that the flight will be delayed is 00.1717 . The probability that it will rain and the flight will be delayed is 00.0505 . What is the probability that it is not raining if the flight leaves on time? Round your answer to the nearest thousandth.\newlineAnswer:\newlineSubmit Answer
  1. Denote Events: Let's denote the events as follows:\newlineR: It will rain.\newlineD: The flight will be delayed.\newlineWe are given the following probabilities:\newlineP(R)=0.19P(R) = 0.19\newlineP(D)=0.17P(D) = 0.17\newlineP(R and D)=0.05P(R \text{ and } D) = 0.05\newlineWe want to find the probability that it is not raining given that the flight leaves on time. This is a conditional probability problem, and we can denote the event of the flight leaving on time as DD'. The probability we are looking for is P(RD)P(R' | D'), where RR' is the complement of R (it does not rain).
  2. Find P(D)P(D'): First, we need to find the probability of the flight leaving on time, which is the complement of the flight being delayed. This is given by:\newlineP(D)=1P(D)P(D') = 1 - P(D)\newlineCalculating this, we get:\newlineP(D)=10.17=0.83P(D') = 1 - 0.17 = 0.83
  3. Find P(R and D)P(R \text{ and } D'): Next, we need to find the probability of it not raining and the flight leaving on time, which is P(R and D)P(R' \text{ and } D'). We can use the fact that P(R and D)P(R \text{ and } D') is the complement of P(R and D)P(R \text{ and } D) within the event DD'. So we have:\newlineP(R and D)=P(D)P(R and D)P(R \text{ and } D') = P(D') - P(R \text{ and } D)\newlineCalculating this, we get:\newlineP(R and D)=0.830.05=0.78P(R \text{ and } D') = 0.83 - 0.05 = 0.78
  4. Calculate P(RD)P(R' | D'): Now, we can find the conditional probability of it not raining given that the flight leaves on time using the formula:\newlineP(RD)=P(R and D)P(D)P(R' | D') = \frac{P(R' \text{ and } D')}{P(D')}\newlineHowever, we need to adjust our previous step because we calculated P(R and D)P(R \text{ and } D'), not P(R and D)P(R' \text{ and } D'). We need to find P(R and D)P(R' \text{ and } D') which is the probability of it not raining and the flight leaving on time. This is given by:\newlineP(R and D)=P(D)P(R and D)P(R' \text{ and } D') = P(D') - P(R \text{ and } D')\newlineBut we already have P(D)P(D'), so we need to find P(R)P(R') first, which is the complement of P(R)P(R):\newlineP(R)=1P(R)P(R') = 1 - P(R)\newlineCalculating this, we get:\newlineP(R)=10.19=0.81P(R') = 1 - 0.19 = 0.81
  5. Adjust Previous Step: Now we can correctly calculate P(R and D)P(R' \text{ and } D') as:\newlineP(R and D)=P(R)×P(D)P(R' \text{ and } D') = P(R') \times P(D')\newlineThis is because RR' and DD' are independent events, and the probability of both occurring is the product of their individual probabilities.\newlineCalculating this, we get:\newlineP(R and D)=0.81×0.83P(R' \text{ and } D') = 0.81 \times 0.83\newlineP(R and D)=0.6723P(R' \text{ and } D') = 0.6723
  6. Calculate P(R and D)P(R' \text{ and } D'): Finally, we can calculate the conditional probability P(RD)P(R' | D') using the correct values:\newlineP(RD)=P(R and D)P(D)P(R' | D') = \frac{P(R' \text{ and } D')}{P(D')}\newlineSubstituting the values we have:\newlineP(RD)=0.67230.83P(R' | D') = \frac{0.6723}{0.83}\newlineCalculating this, we get:\newlineP(RD)0.810P(R' | D') \approx 0.810

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