Probability of choosing 3 specific card out of 100 given 10 tries A deck of $100$ unique cards. 
There are $3$ unique cards that I want to see. 
Each time I draw a card then put it back to the deck after looking at it.
I will then repeat the process until I had done it $10$ times.
What is the probability that after this $10$ draws, I have looked at the $3$ unique cards?
Edit: Chance of drawing each card is $0.01$
 A: Let $p$ denote the probability you want and let $q=1-p$.  It is somewhat easier to compute $q$, the probability that at least one of your favorite three cards is not observed.
To compute $q$:  Note that the probability that a given card is not seen is clearly $\left( \frac {99}{100}\right)^{10}$.  The probability that neither of two (distinct) specified cards is seen is, similarly, $\left( \frac {98}{100}\right)^{10}$.  And the probability that none of your three (distinct) chosen cards is seen is $\left( \frac {97}{100}\right)^{10}$.  By the Principle of Inclusion-Exclusion we get $$q=3\times \left( \frac {99}{100}\right)^{10}-3\times \left( \frac {98}{100}\right)^{10}+\left( \frac {97}{100}\right)^{10}\sim 0.999351931$$
Finally, $$p=1-q\sim \fbox {0.000648069}$$
A: You could work out the following recursion.
Let $p\left(n,k\right)$ denote the probability that after $n$ draws
$k$ of the $3$ specific cards are seen.
Then you are looking for $p\left(10,3\right)$.
Here $p\left(n,k\right)=0$ if $n<k$ and $p(n,0)=1$ and:
$p\left(n,k\right)=\frac{k}{100}p\left(n-1,k-1\right)+\left(1-\frac{k}{100}\right)p\left(n-1,k\right)$.

Another route (suggested by @Lulu)
For $i=1,2,3$ let $E_{i}$ denote the event that card $i$ is not
seen after $10$ draws. Applying inclusion/exclusion and symmetry we find:
$$P\left(E_{1}^{c}\cap E_{2}^{c}\cap E_{3}^{c}\right)=1-P\left(E_{1}\cup E_{2}\cup E_{3}\right)=$$$$1-3P\left(E_{1}\right)+3P\left(E_{1}\cap E_{2}\right)-P\left(E_{1}\cap E_{2}\cap E_{3}\right)$$
That is definitely more handsome than my former suggestion. Credit to @Lulu.
