analysis of a probability question if we have 100 people in total and we are to select each person randomly every day.what is the probability that after 3 years(1095 days) each person has been selected at least once.
for eg. on the first day, the probability will be 1  and second day it will go on decreasinglike  99/100 and so on.
this was actually a puzzle but I wanted to do some probability analysis.
 A: How many functions from $[1,1095]$ to $[1,100]$ are surjective?
Let $N$ be this number then your probability is equal to
$$p=\frac{N}{100^{1095}}=\frac{100!S(1095,100)}{100^{1095}}.$$
Where $S(n,k)$ is a Stirling number of the second kind.
If you are looking for an approximation value of $p$ then for $k$ fixed 
$$k!S(n,k)=\sum_{j=0}^k (-1)^j\binom{k}{j}(k-j)^n\approx k^n-k(k-1)^{n}.$$
and therefore 
$$p\approx \frac{k^n-k(k-1)^{n}}{k^{n}}=1-k(1-1/k)^{n}\approx1-ke^{-n/k}\approx 0.9982442$$
where $k=100$ and $n=1095$. Note that the correct value of $p$ is $0.9983396$.
A: First, you should calcule the probability of the event that $n$-th person is never selected. Let's denote as $A_n$ this event.  
Then, the probability of somebody not being picked any day is 
$$P(A_1) + P(A_2) + \cdots + P(A_{100})$$
What is its complementary event? Such event would have probability,
$$1-\sum^n P(A_n)$$
A: Without replacement everyone gets picked in the first 100 days. With replacement, the probability of any one person being selected (assuming IID) is:
$1-(1-\frac{1}{100})^{1095}$
Since there are 100 people:
$(1-(0.99)^{1095})^{100} = 0.98339722$
