Pirate Scheduling Probability Sparrow is upset. The LWPSA (Large Wooden Pirate Ship Association) recently
passed some new regulations. To ensure that no two pirates are terrorizing the sea at the same time, all pirate ships are required to register ahead of time exactly when they will be patrolling the seas. Sparrow gets frustrated by the new regulations and sends Will to handle registration
in his place.
Will sees there are 30 different possible times to register for (none of these times conflict with
each other). Specifically, each day of the week (Monday to Friday) has 6 available timeslots.
Will doesn’t want the crew to be too overworked, so he decides to select 7 out of the 30 slots
uniformly at random, and register for those.
If all timeslots are equally likely to be chosen (and none of them overlap), what is the probability
that the Ruby Perl (Sparrow’s ship) will end up patrolling the seas at least once each day from
Monday through Friday (inclusive).


I'm really struggling with this question. I was thinking that $|\Omega| = \binom{30}{7}$ and since we know that there are 6 timeslots per day we could do $\binom{6}{1} \times 5$ with one selection leftover. I'm not really sure what to do from here.
 A: Since there has to be a patrol each day, that leaves 2 extra patrols to be distributed on any days. We will think about two cases: 


*

*Both of those patrols occur on the same day. In this case, there are 5 ways to choose which day has 3 patrols, and $\binom{6}{3}=20$ ways to pick patrols that day. The remaining days have $6^4$ possible combinations. Total: $5\cdot 20 \cdot 6^4$

*The duplicate patrols occur on different days. Then there are $\binom{5}{2}=10$ ways to pick the days on which there will be 2 patrols, and for each of those patrols, $\binom{6}{2}=15$ ways to choose the patrols. The remaining patrols have $6^3$ ways of picking patrols. Total: $10 \cdot 15^2 \cdot 6^3$.
Total of two cases: $5\cdot 20 \cdot 6^4+10\cdot15^2\cdot6^3$. Divide that by $\binom{30}{7}$ to get the probability.
A: Hint:  It is easier to compute the chance he doesn't patrol every day.  How many ways are there to patrol without patrolling Monday?  Multiply by five, but notice you have double counted the cases where he doesn't patrol two days.  Think inclusion-exclusion.
A: If all five days are covered, then we have two cases: (a) The two extra patrols fall on the same day, or (b) The two extra patrols fall on different days.
To count the number of ways case (a) could occur, we start with $\binom51$ to choose the day with $3$ patrols, then multiply by $\binom63$ to choose which three patrols, and by $\binom61^4$ to choose single patrols on the other four days.
Can you see how a similar calculation would work for case (b)?
