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A arrives at office at 8-10am regularly; B arrives at 9-11 am every day. Probability that one day B arrives before A? [Assume arrival time of both A and B are uniformly distributed]

A can come between 8 to 10 , which is 120 min duration . A can come on any instant from  this 120 min window .

B can come between 9 to 11 , which is 120 min duration . B also can choose any instant for coming from this 120 min window

so tota possible cases of coming of A nd B =120*120

Now consider the case where B is coming early to A . here common window of 60 min which is from 9 to 10 .

we can choose two time instant from this window by 60C2 ways .

so probability is = 60c2/120*120 = 60*59
--------------=  59/480
2*120*120
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why are you choosing two time instants from 9-10??

and in this question,though the given answer is 1/8 which is very close to your answer but as it is given that data is uniformly distributed,does'nt we have to solve like that?
I choose two instant from 9 -10 because it is the common time window and two instant because one for A's arriving time and one for B's arriving time.  and which value is less we take that for B's arriving time and other one for A's arriving . And it is satisfying the criteria .

And dont find any error in this method and in ans as well .

PS: I dont know the method of uniform distribution
thanks a lot..i got your point

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