Question:
Bob plays a game where, for some number $n$, he chooses a random integer between 0 and $n-1$, inclusive. If Bob plays this game for each of the first four prime numbers, what is the probability that the sum of the numbers he gets is greater than 0?

Answer:
The first four primes are 2, 3, 5, and 7. The only way for the sum of the numbers Bob gets to not be greater than 0 is if he gets a 0 each time he plays the game. The chance that Bob will get a 0 each time he plays is $\frac{1}{2}\cdot \frac{1}{3} \cdot \frac{1}{5} \cdot \frac{1}{7}=\frac{1}{210}$. Therefore, the probability that Bob will not get all 0's is $1-\frac{1}{210}=\boxed{\frac{209}{210}}$.