What is the probability at least one of the balls has the number "1" on it? We have three boxes, the first one contains 5 numbered balls, the second box contains 8 numbered balls and the third box contains 9 numbered balls. 
We choose randomly one ball of each box. 
What is the probability at least one of the balls has the number "1" on it? 
I have done the following: 
Let X be the number of balls with "1" on it. 
Then $P(X\geq 1)=1-P(X<1)=1-P(X=0)$. 
In the first box there are $\binom{4}{1}$ ways to get a ball with "1" on it, in the second box there are $\binom{7}{1}$ ways to get a ball with "1" on it and in the third box there are $\binom{8}{1}$ ways to get a ball with "1" on it, right? 
Then the number of ways to get one ball out of each box is $\binom{5}{1}+\binom{8}{1}+\binom{9}{1}$, or not?   
Does it holds then that $$P(X=0)=\frac{\frac{1}{3}\cdot \binom{4}{1}+\frac{1}{3}\cdot \binom{7}{1}+\frac{1}{3}\cdot \binom{8}{1}}{\binom{5}{1}+\binom{8}{1}+\binom{9}{1}}\quad ?$$ 

The answer is $\frac{17}{45}$.
You can compute the probability of not getting a ball with a $1$ on it and take the complement. For each box with $n$ balls, there are exactly $n - 1$ outcomes that do not result in a $1$. Using independence, you can multiply these probabilities together to find the probability of the event that you obtain no balls with $1$'s on them. Finally, take a complement to find that the answer is $1 - \frac{4}{5} \cdot \frac{7}{8} \cdot \frac{8}{9} = \frac{17}{45}$. 
