A arrives at office at 8- 10 am everyday, B arrives at 9-11 am everyday. Probability B arrives before A? As the title indicated,
A arrives at office at 8- 10 am everyday, B arrives at 9-11 am everyday. Probability that One day B arrives before A? I am bit confused whether to use Poisson (because arrival questions seems to be linked to that) or use Uniform (the question seems uniform) to model that? 
 A: I think you are expected to assume that their arrival times have (continuous) uniform distribution, and are independent. This is not necessarily a reasonable real-world assumption. 
It is convenient, but not necessary, to do a clock change, and assume that A arrives between $0$ and $2$, and B arrives between $1$ and $3$.
Let $X$ be the arrival time of A, and $Y$ the arrival time of B. We want the probability that $Y\lt X$.
Draw the square $S$ with corners $(0,1)$, $(2,1)$, $(2,3)$, and $(0,3)$. Draw the line $y=x$. We want to find the probability that the pair $(X,Y)$ lands in the part of $S$ that is below $y=x$. Call this part $T$. 
Our desired probability is the area of $T$ divided by the area of $S$. The region $T$ is a right triangle with corners $(1,1)$, $(2,1)$, and $(2,2)$. 
So $T$ has area $\frac{1}{2}$. Since $S$ has area $4$, our desired probability is $\frac{1}{8}$.
A: This answer relies on uniform, where it could be done from first principles, and almost no calculations. 
$U_1$ arrival time of $A$. 
$U_2$ arrival time of $B$
$$P(U_1>U_2)=P(U_1>U_2| U_1 >9,U_2<10)P(U_1>9,U_2<10)$$
Uniform conditioned to be in an interval is uniform.  $U_1,U_2$ are independent, so conditioning of one has no affect on the other. 
Therefore, 
$$P(U_1>U_2)=P(U_1'>U_2')P(U_1>9)P(U_2<10).$$
where $U_1',U_2'$ are independent and uniform on $[9,10]$.
Evaluating the three terms on RHS, we obtain: 
$$ \frac 12 \times \frac 12\times \frac 12= \frac 18.$$ 
