There are 28 individuals in a room. Each person will celebrate their birthday in 2016 on the Sunday of the week concluding their birth. Nobody is born on February 29. What is the probability that 2 parties happen in 2016 on the same day.
Note: I asked my lecture to clarify and he said anybody in the last week of December will have their birthday in 2017. This makes the problem so confusing to me.
I thought it might be best to approach it from an inclusion-exclusion perspective? Or could I simply subtract from 1 the probability that nobody shares a birthday? How do I count this? There are 52 Sunday's in 2016, the Sunday after feb 29th only has 5 days preceding it, so will I have to count this one differently as well?
"Or could I simply subtract from $1$ the probability that nobody shares a birthday? How do I count this? There are $52$ Sunday's in $2016$."
First i calculate $P(k)$ the probability that nobody out of $k$ individuals share a birthday week and then its compliment $P(\bar{k})$.
For the sample space $\mathbf{n}(S)$ the number of possible ways $k$ individuals can have birthday weeks is
$$\mathbf{n}(S) = 52^k$$
The number of ways all have different birthday weeks, $\mathbf{n}(B)$, is the number of ways of selecting $k$ different weeks, the number of sequences without repetition.
$$\mathbf{n}(B)={_{52}P_k}=\frac{52!}{(52 - k)!} $$
$$P(k) = \frac{\mathbf{n}(B)}{\mathbf{n}(S)} = \frac{52!}{52^k (52 - k)!}$$
$$\therefore \:\: P(\bar{k}) = 1 - P(k) = 1 - \frac{52!}{52^k (52 - k)!}$$
$$P(\bar{28}) = 1 - \frac{52!}{52^{28} (52 - 28)!} = 1 - \frac{52!}{52^{28}\: 24!}$$