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• 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)?

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1. Submitted By: no-it-all — September 16, 2007 +0 votes   + -

   Constant default probability means it’s always 0.95, regardless of how long you’re twiddling your thumbs on the sidewalk, counting cars.

3. Submitted By: Xavier Cam — September 29, 2007 -2 votes   + -

   .95/3

4. Submitted By: garrincha — October 7, 2007 +11 votes   + -

   Both answers are wrong. The probability of NOT observing the car in 30 minutes is 1-0.95 = 0.05. Since a constant default probability is assumed, the probability of not seeing a car in 10 minutes is (1-0.95)^(1/3), and finally observing a car in 10 minutes we get 1-(1-0.95)^(1/3).

5. Submitted By: Venkatesh — October 14, 2007 not yet rated   + -

   http://discuss.fogcreek.com/techinterview/default.asp?cmd=show&ixPost=713

6. Submitted By: gotta — January 14, 2008 -2 votes   + -

   none of the above answers is right.

   Just think about it, could the possibility of observing a car in 10 mins which is 98.33% according to garrincha’s calculation, be higher than that of observing a car in 30 mins?

   the correct answer is 1-(1-0.95)*3=1-0.15=0.85

7. Submitted By: Chrisping — January 22, 2008 not yet rated   + -

   when you calculated 98.33 you didn’t do it corectly … you pressed (1-0.95)^1 /3 and not (1-0.95) ^(1/3)
   Think about it ?
   I love that explanation JUST think about it - lol

8. Submitted By: kmng2 — February 20, 2008 -2 votes   + -

   0.95. The probabability won’t change because of the time

10. Submitted By: sandesh — March 31, 2008 not yet rated   + -

    in 30 mins the probablity is 0.95
    so 1 min it wud be 0.031
    and for 10 min it wud be 10*0.031

11. Submitted By: mfe08 — April 28, 2008 not yet rated   + -

    Agree with garrincha. Assume the probability of survival in 10 min is p, then probability of suvival for 30 min is p^3 = (1-0.95)
    So the probability of observing a car in 10 mins is
    1-p= 1-(1-0.95)^(1/3)

12. Submitted By: Amit — June 9, 2008 not yet rated   + -

    There are two ways to solve it:

    1)
    If you divide the interval in 3 * 10 min intervals then

    P(A) - Finding Car in 30 min = 0.95
    P(B) - Finding Car in 10 min = X
    P(B’) - Not Finding Car in 10 min = 1 - X

    Given default probability, you can find a Car only in one of the 3 intervals. So:

    P(B) = 0.95 = X * (1 - X) * (1 - X)
    .. solving the above equation you can find the P(B)

    2)
    P(A) - same as above (30 min)
    P(B) - same as above (10 min)

    P(A|B) - Probability of finding a Car in 30 min given that you saw a car in first 10 min.

    P(A|B) = P(A intersect B) / P(B) = 0.95 - given the constant default probability.

    P (A intersect B) = 1/3 - only one of three possibilities

    P(B) = 1 / (3 * 0.95) = 1 / 2.85 = 0.35 - Finding car in 10 min

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