SOLUTION: If you run a red traffic light at an intersection, there is a .1 probability tha tyou will be given a traffic violation. If you runa red traffic light at this intersection five dif
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Question 764344: If you run a red traffic light at an intersection, there is a .1 probability tha tyou will be given a traffic violation. If you runa red traffic light at this intersection five different times, what is the probability of getting at least one traffic violation?
I know the answer is .410, but please explain the process.
Thank you. Found 3 solutions by stanbon, fcabanski, solver91311:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! If you run a red traffic light at an intersection, there is a .10 probability tha tyou will be given a traffic violation. If you runa red traffic light at this intersection five different times, what is the probability of getting at least one traffic violation?
I know the answer is .410, but please explain the process.
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These are independent events.
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Binomial Problem with n = 5 and p(violation) = 0.10 ; P(no vio) = 0.9
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P(at least one in five) = 1 - P(no vios in five) = 1-(0.9)^5 = 0.4095
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Cheers,
Stan H.
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You can put this solution on YOUR website! The probability of getting a ticket at least once is 1 - the probability of never getting a ticket.
The probability of not getting a ticket at each light is .9 (1 - probability of a ticket.) The sum of probabilities of n independent events (if each event probability is A) A^n.
The probability of not getting a ticket at five lights is .9^5.
The probability of getting at least one ticket is 1-.9^5 = 1-.59049 = .40951= approximately .410.
The probability of successes in trials where is the probability of success on any given trial is given by:
Where is the number of combinations of things taken at a time and is calculated by
So in order to find the probability of "at least 1", you would have to know the probablity of exactly 1, plus the probability of exactly 2, plus the probability of exactly 3...and so on.
Which is a lot of nasty arithmetic. Fortunately there is an easier way. Note that the probability of "at least one" is the sum of P(1) + P(2) + P(3) + P(4) + P(5), which are all of the possibilities EXCEPT P(0). Hence if you add P(0) to the above, you must have a result of 1. That's because it is certain that one of the outcomes will occur. That means that if we take the probability of exactly 0 occurances and subtract that from 1, we get the sum of the probabilities of all the other possibilities.
So:
And yet again fortune smiles on us because no matter what might be, and for all real . So really all that needs to be done is:
Just a few quick keystrokes on the calculator.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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