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stellarik [79]
3 years ago
9

Bill booked a ride to the airport. If it rains, the traffic will be bad and the probability that he will miss his flight is 0.06

. If it doesn't rain, the probability that he will miss his flight is 0.01. Suppose the probability that it will rain is 0.21. Suppose also that Bill, unfortunately, missed his flight. What is the probability that it was "actually raining"?
Mathematics
1 answer:
kvasek [131]3 years ago
7 0

Answer:0.614

Step-by-step explanation:

Given

Probability he will miss flight if it rains=0.06

Probability he will miss flight it does not rain=0.01

Given the probability of rain=0.21

Therefore Probability that it will not rain=1-0.21=0.79

Probability that he will miss the flight =0.21\times 0.06+0.79\times 0.01=0.0205

P(actual raining and he missed the flight| he miss the flight)=\frac{0.21\times 0.06}{0.21\times 0.06+0.79\times 0.01}

=\frac{0.0126}{0.0205}=0.614

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Step-by-step explanation:

On the left side, we divide first: 12/4=3

On the left side, we get 3+13, which equals 16.

On the right side, we also divide first: 22/2=11

On the right side, we get 11+2, which equals 13.

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Answer:

Step-by-step explanation:

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b) for n = 3, the probability value from the geometric probability distribution calculator is

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Step-by-step explanation:

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