3. There are 10 multiple-choice questions, each with 4 options. A person randomly selects one option for each question as the answer. If the probability of getting exactly $k$ questions correct is the greatest, then the value of $k$ is $(\quad)$.
(A) 2
(B) 3
(C) 4
(D) 5
3.A.

The answer to each question is randomly and independently selected, and the probability of answering a question correctly is $\frac{1}{4}$, while the probability of answering incorrectly is $\frac{3}{4}$. This is an independent repeated trial, so the probability of answering exactly $k$ questions correctly is
$$
\begin{array}{l}
\mathrm{C}_{10}^{k}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{10-k} . \\
\text { Then }\left\{\begin{array}{l}
\mathrm{C}_{10}^{k}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{10-k} \geqslant \mathrm{C}_{10}^{k-1}\left(\frac{1}{4}\right)^{k-1}\left(\frac{3}{4}\right)^{11-k}, \\
\mathrm{C}_{10}^{k}\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{10-k} \geqslant \mathrm{C}_{10}^{k+1}\left(\frac{1}{4}\right)^{k+1}\left(\frac{3}{4}\right)^{9-k} . \\
\text { Simplifying, we get }\left\{\begin{array}{l}
\frac{1}{k} \geqslant \frac{3}{11-k}, \\
\frac{3}{10-k} \geqslant \frac{1}{k+1} .
\end{array} \text { Solving, we get } \frac{7}{4} \leqslant k \leqslant \frac{11}{4} .\right.
\end{array}\right.
\end{array}
$$

Since $k$ is a positive integer, $k=2$.
Explanation: At this point, we can calculate
$$
P_{10}(2)=C_{10}^{2}\left(\frac{1}{4}\right)^{2}\left(\frac{3}{4}\right)^{8} \approx 0.28 .
$$