simple question on probability.... 
A business man is expecting two telephone calls. Mr. Walia may call any time between $2$ p.m. and $4$ p.m. while Mr. Sharma is equally likely to call any time between $2:30$ p.m. and $3:15$ p.m. The probability that Mr. Walia calls before Mr. Sharma is:

A. $1/18$
B. $1/6$
C. $1/8$
D. None of these

I tried to use the geometrical way to solve the problem by considering a square of $120$ side length and then plotting the graph of $\text{abs}(x-y)=45$... But i am not sure whether i am correct or not?
Any help would be appreciated...thanks in advance...
 A: One simple way of answering this is to note that Mr Walia will call before 2:30 with probability 1/4, and 1/4 is larger than answers a,b, or c.
A: If Mr. Walia calls before 2:30 he will certainly call before Mr. Sharma.   What is the probability that Mr. Walia calls before 2:30?
If Mr. Walia calls after 3:15 he will certainly not call before Mr. Sharma.  
What is the probability of Mr. Walia calling between 2:30 and 3:15?   If Mr. Walia does so, then what is the conditional probability that he calls before Mr. Sharma?
Put it together.

$$\mathsf P\big(W \leq S\big) ~=~ \mathsf P\big(W < 2{:}30\big)+\mathsf P\big(2{:}30\leq W\leq 3{:}15\big)~\mathsf P\big(W<S~\big\vert~ 2{:}30\leq W\leq 3{:}15\big)$$
A: Let $x$ and $y$ denote the times when Mr. Walia and Mr. Sharma call, respectively. For convenience let us express these times in number of minutes past 2:00 pm. Therefore $x$ is uniformly distributed in the interval $[0,120]$ and $y$ is uniformly distributed in $[30, 75]$.
Mr. Walia calls before Mr. Sharma if and only if $x < y$.
Although it is not stated in the problem, I assume that $x$ and $y$ are independent. Therefore, the joint distribution of $x$ and $y$ is uniform in a rectangular region with $x \in [0,120]$ and $y \in [30,75]$. More explicitly, the area of this region is $120 \cdot (75 - 30) = 5400$, and so the joint pdf of $x$ and $y$ is
$$p(x,y) = \begin{cases}
\frac{1}{5400} & \text{if }x \in [0,120] \text{ and } y \in [30,75] \\
0 & \text{otherwise}
\end{cases}$$
The subregion where $x < y$ is the region bounded by $x=0$, $y = 30$, $y = 75$, and $x=y$. This decomposes into a rectangle with area $30\cdot 45 = 1350$ and a triangle with area $\frac{1}{2}\cdot 45 \cdot 45 = 1012.5$, so the total area of the region is $2362.5$. Then the probability that $(x,y)$ lies in this region is $2362.5 / 5400 = 0.4375 = 7/16$.
