There are $9$ new tennis balls in the box. Every game there will be randomly chosen $3$ balls. After the game, these $3$ balls are used. These used balls will put back in the box after every game. What is the probability that after $3$ games there is at least $1$ new tennis ball in the box?
My solution:
$A$ - after $3$ games at least $1$ ball not used
$A_i$ - for game number $3$ new balls $i=1,2,3$
$A=A_1A_2A_3$
This is quite a simple problem once you realise you can 'flip the question on its head'. So rather than considering the chance of having at least $1$ new tennis ball after $3$ games, let's instead consider the chance that there are no new tennis balls after $3$ games. This turns out to be a very simple calculation because there is only one way in which this can happen: no ball can be used twice. If any ball is used twice, then there will be at least one new tennis ball after $3$ games.
Now that we have thought carefully about the problem, the actual mathematics is very easy:
$$ P(\text{Never using the same ball twice}) = \frac{6}{9} \times \frac{5}{8} \times \frac{4}{7} \times \frac{3}{9} \times \frac{2}{8} \times \frac{1}{7} = \frac{5}{1764} $$
Hence,
$$ P(\text{Using the same ball twice})=\frac{1759}{1764} $$
as the probabilities have to add up to $1$.
And, as we concluded earlier, if you use a single ball more than once, then there will be at least one new ball after $3$ games.