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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 0.05 . The probability that it will rain and the flight will be delayed is 0.03 . What is the probability that it is 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 $0$ . $08$ and the probability that the flight will be delayed is $0$ . $05$ . The probability that it will rain and the flight will be delayed is $0$ . $03$ . What is the probability that it is raining if the flight leaves on time? Round your answer to the nearest thousandth. $\newline$ Answer: $\newline$

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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

.

05

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

0

.

03

. What is the probability that it is raining if the flight leaves on time? Round your answer to the nearest thousandth.

\newline

Answer:

\newline

Events Denoted: Let's denote the events as follows: $\newline$ R: It will rain. $\newline$ D: The flight will be delayed. $\newline$ L: The flight leaves on time (which is the complement of the flight being delayed). $\newline$ We are given the following probabilities: $\newline$ $P(R) = 0.08$ (Probability that it will rain) $\newline$ $P(D) = 0.05$ (Probability that the flight will be delayed) $\newline$ $P(R \text{ and } D) = 0.03$ (Probability that it will rain and the flight will be delayed) $\newline$ We need to find the probability that it is raining given that the flight leaves on time, which can be written as $P(R | L)$ . According to the definition of conditional probability, we have: $\newline$ $P(R | L) = \frac{P(R \text{ and } L)}{P(L)}$ $\newline$ Since L is the complement of D, we can say $P(L) = 1 - P(D)$ . We also know that $P(R \text{ and } L) = P(R) - P(R \text{ and } D)$ , because the probability that it is raining and the flight leaves on time is the probability that it is raining minus the probability that it is raining and the flight is delayed. $\newline$ Let's calculate $P(L)$ first: $\newline$ $P(L) = 1 - P(D)$ $\newline$ $P(L) = 1 - 0.05$ $\newline$ $P(D) = 0.05$ $0$
Calculate $P(L)$ : Now let's calculate $P(R \text{ and } L)$ :
$P(R \text{ and } L) = P(R) - P(R \text{ and } D)$
$P(R \text{ and } L) = 0.08 - 0.03$
$P(R \text{ and } L) = 0.05$
Calculate $P(R \text{ and } L)$ : Finally, we can calculate $P(R | L)$ : $P(R | L) = \frac{P(R \text{ and } L)}{P(L)}$ $P(R | L) = \frac{0.05}{0.95}$ $P(R | L) \approx 0.05263$ Rounding to the nearest thousandth, we get: $P(R | L) \approx 0.053$

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