There are 28 people, none of whom are born on the 29th of Feb. Each individual organises a birthday party on the Sunday concluding the week of their birthday. What is the probability that at least 2 individuals have their party on the same day in 2016?
Ok so I really have no idea how to tackle this problem as I have seen nothing like it in any of the course notes. I feel like it could be related to the topic of inclusion-exclusion though. I have worked out that there are exactly 52 Sundays in 2016. This means that the probability of a particular student having a birthday on any particular Sunday is 1 in 52. The probability that a birthday party is held on a particular Sunday is 28 in 52. But I am not sure how to proceed from here, any help would be greatly appreciated.
I'd first like to address the "edge effects," as not every Sunday has an equal probability of being selected as a birthday for any random day. Notice that for birthdays between (and including) January 1 and 3 (3 days), the party will fall on January 3. For birthdays between (and including) December 28 and 31 (4 days), the party would be held on January 3 as well. Notice that if a birthday is on December 26 or 27, no birthday party will be held in the year 2016. For consistency's sake, since we disallow any leap year birthdays, let us also disallow Dec. 26 and 27.
Notice that every Sunday except for March 6 (leap day's Sunday) will have $7$ "covered dates." March 6 will have only $6.$ The sample space is $363$ dates, of which we select $28.$ This can be done in $\dbinom{363}{28}$ ways.
We will now use complementary counting to find our answer. Of course, you can use principle of inclusion / exclusion, but the counting is far more tedious. If no two people have the same Sunday as their party dates, then we have to select $28$ distinct Sundays. We do this by casework with two cases: 1) March 6 is an included Sunday, and 2) March 6 is not an included Sunday.
Case 1: Besides March 6, there are $51$ other Sundays in 2016, as you have mentioned. We choose $27$ of these, then select the specific day (out of the "covered" $7$) for every one. We then choose one day out of $6$ for the March 6 birthday. Our count becomes $$A = \dbinom{51}{27} \times 7^{27} \times 6.$$
Case 2: We have $51$ Sundays from which to pick $28.$ We then select the specific day for each. This counts $$B = \dbinom{51}{28} \times 7^{28}.$$
Notice, by the way, that $A = B.$
We have our complementary count, so all we have to do now is take $$1 - \frac{A + B}{\dbinom{363}{28}}.$$ This is approximately $\boxed{0.9996622},$ very close to $1.$ And we are done.