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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