cars on the road 3/5 or 60%
Pankaj Gupta
Hi
Paulius
I do agree with Paulius.
Pe Woerner
No I think, pankaj's solution is flawless.
vivek gupta
1-pr(not seeing in ALL intervals) = pr(seeing cars in every 5.min interval). I think this is different from seeing car at one of 5 min. intevals in 20 min. period
Paulius
If we go by your formula , then
vivek gupta
true...
shailesh kumar
P(see a car in 20 minutes) = 609/625
Adrian Gilby
Damn. Forgot to say ... so Pankaj is right.
Adrian Gilby
Ok, let's modify probability of apearing car in 20 min. period and let it be equal to 1. Now by calculating probability using method i proposed we are getting 1/4, by calculating probability using method Gupta probosed we are getting 1.
Paulius
Yup, and Pankaj and Vivek are correct. Look at it like this -- if you're absolutely guaranteed to see a car in _any_ 20 minute period, then you must be seeing a car _constantly_.
Adrian Gilby
Paulius,
A.T.
Probability c = 609/625 we can treat like we did first 625 observations and found that 609 observations were successfull (car appeared in 20 min. period). How many observations we can do in 5 min. intervals during same time ? I think 2500 while cars were comming same way.
Paulius
Vin
I just stumbled across this forum and I though I'd mention another way to solve this problem, one that incidentally doesn't rely on the numbers being so judiciously chosen (but on the other hand does require a calculator.)
Carter Shanklin
Pankaj's explanation of how he arrives at the solution is, in my opinion, the most direct --and therefore the most simple-to-understand. Good job :)
Jim Tran
just wanted to point out that the main reason that
Cheradenine Zakalwe
Pankaj is correct. Sometimes.
J. Martin
The 5-minute analogue of the statement 'The probability is 609/625 of seeing one and only one car in 20 minutes' is 'The probability is x of seeing one and only one car in 5 minutes.'
rob mayoff
(Note to anybody not reading the full discussion: I agree with Pankaj. This a tangent to some degree. Please read my post above.)
J. Martin
I'm no mathematician, but I really don't get Andrian's idea that if the probability of seeing a car in any 20 minute time period is 1, then the probability of seeing a car in any 5 minute time period is also 1.
Marko
Poisson is to probability what the straight line is to calculus. It should be the first tool you turn to for a quick and dirty calculation. It is not "something to avoid because it is too complicated"
MFT
So basically, no need to ask this question in an interview!
Muhammad Omer Iqbal
p =3/5 is the correct answer.
Dmitriy Furer
Paulius is wrong. The probability of seeing a car in 5 minutes is NOT p/4 (where p=probability of seeing a car in 20 minutes). If that was correct, then you could also argue that the probability of seeing a car in 40 minutes would be.. what, p*2 ? That would be p >=1 which as we all know is impossible.
Guillermo Rodriguez
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