Solution:

The answer is (D). Indeed, let us compute the probability $p$ that Matteo gets all the answers wrong (the required probability is then $1-p$): for the first question there is one wrong answer out of two total answers, so Matteo has probability $1/2$ of getting it wrong. For the second question the incorrect options are two out of three, so Matteo has probability $2/3$ of getting it wrong. Similarly, for question number $j$, there are $j$ wrong options out of a total of $j+1$, so filling in the test at random Matteo has probability $\frac{j}{j+1}$ of getting it wrong. The probability of getting all the answers wrong is the product of the probabilities of getting each of the answers wrong, so
$$p = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot \ldots \cdot \frac{11}{12}$$
Noting that each denominator cancels with the numerator of the following fraction, we immediately get $p = \frac{1}{12}$, so the required probability is $1-p = \frac{11}{12}$.