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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]

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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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U1 arrival time of A. U2 arrival time of B

Red area shows A’s distribution and Blue area shows B’s distribution.

P(U1>U2) = P(U1>U2 | U1>9, U2<10 ) P(U1 > 9, U2<10 ) 

Uniform conditioned to be in an interval is uniform. U1,U2 are independent, so conditioning of one has no affect on the other.

Therefore,

P(U1>U2) = P( U′1 > U′2 ) P(U1 > 9 ) P( U2 < 10 )

where U′1,U′2 are independent and uniform on [9,10]. (how i interpret this is, the probability of anything happening in this area(where red and blue intersect) is ½ which includes U’1 > U’2 since it is uniformly distributed)

P( U′1 > U′2 ) = ½

P(U1 > 9 ) = (10-9)*(1/2) = ½

P(U1 > 9 ) = (11-10)*(1/2) = ½

Evaluating the three terms, we obtain:

1/2×1/2×1/2=1/8.

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