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At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.1 and the probability that the flight will be delayed is 0.16. The probability that it will rain and the flight will be delayed is 0.02. What is the probability that it is not raining and the flight leaves on time? Round your answer to the nearest thousandth.

Question

At LaGuardia Airport for a certain nightly flight, the probability that it will rain is 0.1 and the probability that the flight will be delayed is 0.16. The probability that it will rain and the flight will be delayed is 0.02. What is the probability that it is not raining and the flight leaves on time? Round your answer to the nearest thousandth.

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Solution

Let's denote the following events:

  • R R : It will rain.
  • D D : The flight will be delayed.

We are given:

  • P(R)=0.1 P(R) = 0.1
  • P(D)=0.16 P(D) = 0.16
  • P(RD)=0.02 P(R \cap D) = 0.02

We need to find the probability that it is not raining and the flight leaves on time. This can be expressed as P(¬R¬D) P(\neg R \cap \neg D) .

First, we use the formula for the complement of an event: P(¬R)=1P(R) P(\neg R) = 1 - P(R) P(¬R)=10.1=0.9 P(\neg R) = 1 - 0.1 = 0.9

Similarly, P(¬D)=1P(D) P(\neg D) = 1 - P(D) P(¬D)=10.16=0.84 P(\neg D) = 1 - 0.16 = 0.84

Next, we use the principle of inclusion-exclusion for the union of two events: P(RD)=P(R)+P(D)P(RD) P(R \cup D) = P(R) + P(D) - P(R \cap D) P(RD)=0.1+0.160.02=0.24 P(R \cup D) = 0.1 + 0.16 - 0.02 = 0.24

The probability that it is not raining and the flight leaves on time is the complement of the union of R R and D D : P(¬R¬D)=1P(RD) P(\neg R \cap \neg D) = 1 - P(R \cup D) P(¬R¬D)=10.24=0.76 P(\neg R \cap \neg D) = 1 - 0.24 = 0.76

Therefore, the probability that it is not raining and the flight leaves on time is 0.76 0.76 . Rounding to the nearest thousandth, the answer is 0.760 0.760 .

This problem has been solved

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