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At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.08 and the probability that the flight will be delayed is 0.12 . The probability that it will not rain and the flight will leave on time is 0.84 . What is the probability that it is raining and the flight is delayed? 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 $0$ . $08$ and the probability that the flight will be delayed is $0$ . $12$ . The probability that it will not rain and the flight will leave on time is $0$ . $84$ . What is the probability that it is raining and the flight is delayed? 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

.

08

and the probability that the flight will be delayed is

0

.

12

. The probability that it will not rain and the flight will leave on time is

0

.

84

. What is the probability that it is raining and the flight is delayed? Round your answer to the nearest thousandth.

\newline

Answer:

Events Denotation: Let's denote the events as follows: $\newline$ R: It will rain. $\newline$ D: The flight will be delayed. $\newline$ N: It will not rain and the flight will leave on time. $\newline$ We are given the following probabilities: $\newline$ $P(R) = 0.08$ $\newline$ $P(D) = 0.12$ $\newline$ $P(N) = 0.84$ $\newline$ We need to find the probability that it is raining and the flight is delayed, which is $P(R \text{ and } D)$ .
Find $P(R \text{ or } D)$ : First, we need to find the probability that it will rain or the flight will be delayed, which is $P(R \text{ or } D)$ . This can be found using the complement of the probability that it will not rain and the flight will leave on time, which is $P(N)$ . $\newline$ Since $P(N)$ is the probability that it will not rain and the flight will leave on time, the complement is the probability that it will rain or the flight will be delayed, so: $\newline$ $P(R \text{ or } D) = 1 - P(N)$ $\newline$ $P(R \text{ or } D) = 1 - 0.84$ $\newline$ $P(R \text{ or } D) = 0.16$
Use Inclusion-Exclusion Principle: Now, we can use the Inclusion-Exclusion Principle to find $P(R \text{ and } D)$ . The principle states: $\newline$ $P(R \text{ or } D) = P(R) + P(D) - P(R \text{ and } D)$ $\newline$ We can rearrange this formula to solve for $P(R \text{ and } D)$ : $\newline$ $P(R \text{ and } D) = P(R) + P(D) - P(R \text{ or } D)$ $\newline$ Substituting the values we have: $\newline$ $P(R \text{ and } D) = 0.08 + 0.12 - 0.16$ $\newline$ $P(R \text{ and } D) = 0.20 - 0.16$ $\newline$ $P(R \text{ and } D) = 0.04$

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