Answer
See below.
Work Step by Step
We can get this by using the fact that the result is $1-P(\text{no people have birthdays on the same day}).$
We know that probability$=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}.$ There are $13$ people in total. The number of all outcomes is $365^{35}$, because all people can have birthdays on each day.
If we want to choose $k$ elements out of $n$ regarding the order, not allowing repetition, we can do this in $_{n}P_k=\frac{n!}{(n-k)!}$ ways.
The order matters here when choosing the birthdays, thus we have to use permutations. We have $35$ people for $365$ spots, thus the number of good outcomes is: $_{365}P_{35}=\frac{365!}{(365-35)!}$
Thus probability=$1-\frac{\frac{365!}{(365-35)!}}{365^{35}}\approx0.814$
