If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?
1st Way
Let p is a probability to see a car in 10 minutes.
Then (1-p) is probability NOT to see a car in 10 minutes.
Then probability NOT to see a car in 30 minutes is (1-p)*(1-p)*(1-p).
(1-p)^3 == 0.05
So p = 1-0.05^(1/3)~ 0.63
2nd Way
You have to look at your probability of NOT seeing a car, which is .05 in 30 minutes. In order to break this down into 10 minute chunks, you need to figure out how you arrived at that probability, which would be x * x * x = .05, so x ^ 3 = .05.
2 comments:
Why intervals of 10 min and not 15 min?
because question is asked for 10 min. right ?
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