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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 0.15 . The probability that it will rain and the flight will be delayed is 0.01 . What is the probability that the flight would leave on time when it is not raining? Round your answer to the nearest thousandth.
Answer:

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 $0$ . $15$ . The probability that it will rain and the flight will be delayed is $0$ . $01$ . What is the probability that the flight would leave on time when it is not raining? Round your answer to the nearest thousandth. $\newline$ Answer:

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Find probabilities using the addition rule

Full solution

Q. 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

.

15

. The probability that it will rain and the flight will be delayed is

0

.

01

. What is the probability that the flight would leave on time when it is not raining? Round your answer to the nearest thousandth.

\newline

Answer:

Calculate $P(R')$ : Let's denote the events as follows: $\newline$ $R$ : It will rain. $\newline$ $D$ : The flight will be delayed. $\newline$ $R'$ : It will not rain (the complement of $R$ ). $\newline$ $D'$ : The flight will leave on time (the complement of $D$ ). $\newline$ We are given the following probabilities: $\newline$ $P(R) = 0.16$ $\newline$ $P(D) = 0.15$ $\newline$ $P(R \text{ and } D) = 0.01$ $\newline$ We want to find $R$ $0$ , which is the probability that the flight leaves on time and it is not raining. $\newline$ First, we need to find $P(R')$ , which is the probability that it does not rain. This is the complement of $R$ $2$ , so we calculate it as: $\newline$ $R$ $3$ $\newline$ $R$ $4$ $\newline$ $R$ $5$
Find $P(D')$ : Next, we need to find $P(D)$ , which is already given as $0.15$ . To find $P(D')$ , we calculate the complement of $P(D)$ : $\newline$ $P(D') = 1 - P(D)$ $\newline$ $P(D') = 1 - 0.15$ $\newline$ $P(D') = 0.85$
Use Probability Rules: Now, we need to find $P(D' \text{ and } R')$ , but we don't have this probability directly. We can use the fact that $P(D' \text{ and } R')$ is related to $P(D \text{ and } R)$ , $P(D')$ , and $P(R')$ . $\newline$ We know that $P(D \text{ and } R) + P(D' \text{ and } R') = P(R')$ , because the events $(D \text{ and } R)$ and $(D' \text{ and } R')$ are mutually exclusive and their union is the event $R'$ . $\newline$ So we can write: $\newline$ $P(D' \text{ and } R') = P(R') - P(D \text{ and } R)$ $\newline$ $P(D' \text{ and } R')$ $0$ $\newline$ $P(D' \text{ and } R')$ $1$

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