When will Andrea arrive before Bert? The question was as follows-

on any given day, Andrea is equally likely to clock in at work any time from 8:50am to 9:06am. Similarly, Bert is equally likely to to clock in at work at any time from 8:55am to 9:10am.
If the time clock at work is digital and shows only hours and minutes, what fraction of the time will Andrea arrive before Bert? They are considered to have arrived at the same time if the hours and minutes are the same. If they are not consider them arriving at different times.

I came up with $97/136$  as my final result, but I saw another source saying the answer was $7/9$. Can you please explain how it is done? Thanks.
 A: Let me try to solve simply.
There are $17\cdot16 = 272$ possible combinations of clock times (points in time) for Andrea & Bert 
There are $12\cdot12 = 144$ possible points, when both can be present, and Andrea can only be later in this range.
Of these, on $12$ , they will arrive simultaneously, and Andrea will be later on half of the remaining $132$, thus not earlier on  $78$ of the total points
Hence Pr = $\dfrac{(272-78)}{272} = \dfrac {97}{136}$
Your answer is right !   
A: Your result is okay.
Let $E$ denote the event that Andrea and Bert both arrive at a time such that the other can arrive at the same time (so from 8:55 to 9:06 wich are $12$ arrival times).
Let $A$ denote the arrival time of Andrea (she has $17$ possible arrival times) and let $B$ denote the arrival time of Bert (he has $16$ possible arrival times).
It is handsome to look at the probability that Bert will not arrive later than Andrea. This because this can only happen if event $E$ also happens.
$P\left(\left\{ B\leq A\right\} \right)=P\left(E\cap\left\{ B\leq A\right\} \right)=P\left(B\leq A\mid E\right)P\left(E\right)$
Then $P\left(E\right)=\frac{12}{17}\frac{12}{16}$ and $P\left(B=A\mid E\right)=\frac{1}{12}$.
Also we have the equations: $$P\left(B\geq A\mid E\right)+P\left(B\leq A\mid E\right)=1+P\left(B=A\mid E\right)=1+\frac1{12}$$
and based on symmetry: $$P\left(B\geq A\mid E\right)=P\left(B\leq A\mid E\right)$$
Based on that we find $P\left(B\leq A\mid E\right)=\frac{13}{24}$ hence $P\left(\left\{ B\leq A\right\} \right)=\frac{13}{24}\frac{12}{17}\frac{12}{16}$.
Finally $P\left(\left\{ A<B\right\} \right)=1-\frac{13}{24}\frac{12}{17}\frac{12}{16}=\frac{97}{136}$.
A: You can always use brute force in a case like this, but I'm sure there's more elegant solutions.  Andrea has 17 possible clock in times,  whereas Bert has 16 possible clock in times, everything is equally possible and independant, so there are $16\cdot 17=272$ possible combinations.    Now,  we're only interested in the ones where Andrea is first.   Well,  if she clocks in from 8:50 to 8:54, she's first in all 16 possible clock in times for Bert.   So that's $5\cdot 17=85$ possibilities.   Now,  each minute we add to Andrea's clock in time, we remove one clock in time from Bert's possibilities for Andrea to be in first:  So at 8:55 we have 15 possibilities,  8:56 we have 14 possibilities, all the way down to 9:06 where we have 4 possibilities.   So this is the sum of the numbers from 4 to 15.   Doing this as the sum of numbers from 1-15 as $(15)(16)/2=120$, then taking away 6 for the numbers 1-3, we get 114 possibilites, to add to the 85we had to begin with.  so that's $85+114=199$, so the probability will be $199/272$.   At least if I didn't make any fencepost errors, it's late :)
A: Let's see the time where Bert arrives and calculate the probability that, if Bert arrives at that time, Andrea arrives first.
Let $\mathcal X$ be the time Bert arrives at.
Let $\mathscr P_{\mathcal X}$ be the probablility that Andrea arrives before $\mathcal X$. We can calculate it just by counting the number of possibilities (s)he has to be the first versus the number of possibilities (s)he has to arrive at all (this last one is $17$)
If $9:07 \le \mathcal X \le 9:10$, then $\mathscr P_{\mathcal X} = 1$
If $\mathcal X = 9:06$, then $\mathscr P_{\mathcal X} = \frac{16}{17}$
If $\mathcal X = 9:05$, then $\mathscr P_{\mathcal X} = \frac{15}{17}$
...
If $\mathcal X = 8:55$, then $\mathscr P_{\mathcal X} = \frac{5}{17}$  
We now have a probability for the $16$ possibilities of arrival for Bert.
The global probability If $\mathscr P$ that Andrea is first is therefore :
$$
\mathscr P = \frac{1}{16}\sum_{\mathcal X \in [8:55~;~9:10]}{(\mathscr P_{\mathcal X})}= \frac{\frac{5+...+16}{17}+4}{16}= \frac{\frac{6\times21}{17}+4}{16}=\frac{194}{272}
$$
... which is not far from $\frac79$, but not quite that still. ($\frac79 \approx 0.7777... \sim 0.7132... \approx \frac{194}{272}$)
I know this is not elegant but this should work anyway
