To determine the probability of your professor selecting 3 questions out of the 5, you can use the concept of combinations. The number of ways to choose 3 questions from 5 is given by:

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$

So, there are 10 distinct sets of 3 questions your professor can select from the 5.

Should You Study 1 Question or All of Them?

If you study only 1 question, the probability of it being among the 3 selected is given by:

$P(\text{1 question studied is selected}) = \frac{\binom{1}{1} \cdot \binom{4}{2}}{\binom{5}{3}}$

Breaking it down:

• $\binom{1}{1} = 1$ (ways to pick the studied question),
• $\binom{4}{2} = 6$ (ways to pick 2 other questions from the remaining 4),
• $\binom{5}{3} = 10$ (total possible sets of 3 questions).

P(\text{1 question studied is selected}) = \frac{1 \cdot 6}{10} = 0.6 \text{ (60% chance)}.

If you study all 5 questions, you are guaranteed to cover all possible outcomes, which gives you a 100% chance of being prepared.

---

Recommendation

If you want to maximize your chances, it is safer to study all 5 questions. However, if time is a constraint, focusing on 3 or 4 questions may still provide a high probability of covering the selected questions.

---

Would you like me to help you analyze this further or break down your preparation strategy? Here are 5 related questions for exploration:

1. What if the professor picks a different number of questions (e.g., 2 or 4)? How does this change the probability?
2. What’s the chance of guessing all 3 questions correctly if you study only 2?
3. How does studying more questions impact preparation efficiency versus probability coverage?
4. Can we model this situation using probabilities to find an optimal strategy?
5. What other mathematical models can help in decision-making for limited preparation time?

Tip: Diversify your preparation to cover a mix of likely and unlikely questions—it maximizes both confidence and coverage.