6. An urn contains $N$ balls $(N>3)$ numbered from 1 to $N$. If two balls with numbers that are not multiples of 3 and one ball with a number that is a multiple of 3 are removed from the urn, the probability of drawing a multiple of 3 is less than it was with the full urn. What can be deduced about $N$?

(A) $N$ is certainly a multiple of 3

(B) $N$ is not a multiple of 3

(C) $N$ is certainly odd

(D) $N$ is certainly even

(E) none of the previous statements can be deduced.
6. The answer is (B). In the presence of a finite number of possible and equally likely outcomes of an experiment (in our case, the extraction of a ball from the urn), the probability of an event (for us, the extraction of a ball with a number that is a multiple of 3) is the ratio between the number of favorable outcomes and the number of possible outcomes.

Let $k$ be the number of balls with a number that is a multiple of 3 initially in the urn; we will have $3 k \leq N \leq 3 k+2$. After removing 3 balls from the urn, it will contain $N-3$ balls, of which $k-1$ have a number that is a multiple of 3. Let $p_{1}$ and $p_{2}$ be the probabilities of drawing a ball with a number that is a multiple of 3 before and after the removal of the 3 balls, respectively. Then:

$$
p_{1}-p_{2}=\frac{k}{N}-\frac{k-1}{N-3}=\frac{N-3 k}{N(N-3)}
$$

In the case where $N=3 k$, we have $p_{1}=p_{2}$ (so statement $\mathbf{A}$ is false); if instead $N=3 k+1$ or $N=3 k+2$, the numerator is positive, so the two numbers $3 k+1$ and $3 k+2$, being consecutive integers, are necessarily one even and one odd, which means we cannot definitively state that $N$ is even or that $N$ is odd, thus excluding answers $\mathbf{D}$ and $\mathbf{C}$.