Quant Probability Parking Question

There are 10 parking spots in front of Quant Essential headquarters. By 9:00 AM, all 10 parking spots are occupied. Quant Essential's 3 executives head home early in the afternoon, vacating their parking spots at random times distributed independently and uniformly between 12:00 PM and 3:00 PM. Andy is one of Quant Essential's 9 lazy employees; lazy employees arrive at the office at random times distributed independently and uniformly between 12:00 PM and 3:00 PM. If a parking spot is vacant when a lazy employee arrives, then that employee will occupy that spot until 5:00 PM. Otherwise, the lazy employee will call in sick and return home. What is the probability that Andy calls in sick?

The hint asks us to only consider the order of arrivals/exits since all the entry and exit times are iid. I began to solve by finding the complement of the answer first.

So basically Andy won't go home if we sum these cases

However, the complement of this is not getting me the answer. What am I doing wrong? Are we not supposed to break into cases at all?

My attempt is that Andy can park when

which seems to add up to $\frac{581}{1980}$,

so the probability Andy calls in sick is $\frac{1399}{1980} = 0.70\overline{65}$.

Let Executives be E and Lazies be L.

Hint/Idea: Find the expected number of L that worked. By linearity of expectation, the probability that Andy worked is $ \frac{1}{9}$ of that value.

Consider the last person that arrived, and move down that sequence till you know for certain how many L worked. For example, if the last 3 to arrive are LLL, then we know that 3 L worked that day (not necessarily the last 3 to arrive). Conversely, if the last 3 to arrive are EEE, then we know that 0 L worked that day.

Hence the expected number of L that work is $ \frac{ 581}{ 220}$

And the probability that Andy works is $ \frac{ 581} { 1980}$, and the probablity that Andy calls in sick is $ \frac{ 1399}{1980}$.