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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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B can arrive prior to A in the interval of $9-10$ Am.

In this graph i have changed timings,  that A arrives between $0$ and $2$, and B arrives between $1$ and $3$.

Now we are only interested in area where B arrives prior to A, this will be the area where $(x,y)$ where $x > y$.

Because x which is arrival time of A must be greater than that of B's $(y)$.

Shaded area is the area where $x > y$,

Area of triangle = $\large \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Area of sample space (square) = $4$

Probability of B's arrival prior to A's = $\Large \frac{\frac{1}{2}}{4} = \frac{1}{8}$

 

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