Example 4: Mike participates in a ten-question multiple-choice test, getting one point for each correct answer, losing one point for each wrong answer, and getting zero points for unanswered questions. His goal is to score at least 7 points, with 7 points being the passing score. Mike is now certain that his answers to the first six questions are correct, and the probability of getting each of the remaining questions right is $p(0<p<1)$. How many more questions should Mike answer to maximize his probability of passing?
【Analysis】Mick got six questions right.
The probability of passing by doing one more question is $P_{1}=p$, the probability of passing by doing two more questions is $P_{2}=p^{2}$, the probability of passing by doing three more questions is
$$
P_{3}=p^{3}+\mathrm{C}_{3}^{2} p^{2}(1-p)=p^{2}(3-2 p) \text {, }
$$

The probability of passing by doing four more questions is
$$
P_{4}=p^{4}+\mathrm{C}_{4}^{3} p^{3}(1-p)=p^{3}(4-3 p) \text {. }
$$

Obviously, $P_{1}>P_{2}, P_{3}>P_{4}$.
Therefore, we only need to compare the size of $P_{1}$ and $P_{3}$.
When $P_{1}p$, the inequality simplifies to
$$
p^{2}+3 p(1-p)>1 \text { or }(1-p)(2 p-1)>0 \text {. }
$$

Solving this, we get $\frac{1}{2}P_{3}$, in this case, the probability of passing by answering seven questions is the highest;

When $p=\frac{1}{2}$, $P_{1}=P_{3}$, in this case, the probability of passing by answering seven questions and answering nine questions is the highest.