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Let's go through each part of the problem one by one.

### Part A
For a single question with 5 possible answers, the probability of answering correctly at random is \( \frac{1}{5} \).

So, the answer to part (A) is:
\[
\frac{1}{5}
\]

---

### Part B
For part (B), we want to find the probability that the student answers no questions correctly out of 19 questions.

Since each question has a \( \frac{1}{5} \) probability of being correct, the probability of getting a question wrong is \( 1 - \frac{1}{5} = \frac{4}{5} \).

Let \( X \) be the number of questions answered correctly. \( X \) follows a **Binomial distribution** with parameters \( n = 19 \) and \( p = \frac{1}{5} \).

The probability of answering no questions correctly (i.e., \( X = 0 \)) is:
\[
P(X = 0) = \left(\frac{4}{5}\right)^{19}
\]

Calculating this value:
\[
\left(\frac{4}{5}\right)^{19} \approx 0.0132
\]

So, the answer to part (B), rounded to four decimal places, is:
\[
0.0132
\]

---

### Part C
For part (C), we need to find the probability that the student answers exactly one question correctly out of 19.

Again, using the **Binomial distribution** with \( n = 19 \) and \( p = \frac{1}{5} \), the probability of answering exactly one question correctly (i.e., \( X = 1 \)) is given by:
\[
P(X = 1) = \binom{19}{1} \left(\frac{1}{5}\right)^1 \left(\frac{4}{5}\right)^{18}
\]

Calculating each component:
- \( \binom{19}{1} = 19 \)
- \( \left(\frac{1}{5}\right)^1 = 0.2 \)
- \( \left(\frac{4}{5}\right)^{18} \approx 0.0174 \)

Putting it all together:
\[
P(X = 1) = 19 \times 0.2 \times 0.0174 \approx 0.0661
\]

So, the answer to part (C), rounded to four decimal places, is:
\[
0.0661
\]

---

### Summary of Answers
- **Part A:** \( \frac{1}{5} \)
- **Part B:** \( 0.0132 \)
- **Part C:** \( 0.0661 \)

Would you like further explanations on any part of this problem, or have any specific questions?

---

### Related Questions
1. What would the probability be if there were 25 questions instead of 19?
2. How does the probability of no correct answers change if each question has only 4 possible answers?
3. What is the expected number of correct answers for this test?
4. How do we calculate the standard deviation of correct answers in this scenario?
5. What is the probability that the student answers at least one question correctly?

---

### Tip
When dealing with probabilities in multiple-choice questions, using the Binomial distribution is very useful for calculating scenarios like "exactly" or "at least" a certain number of correct answers.

A student is given a test consisting of 19 multiple-choice questions, each having 5 possible answers. The student didn't study, so he decides to answer every question at random.

A. For one question, what is the probability that a student will answer it correctly if he has 5 possible answers and just answers it at random?

B. What is the probability that out of 19 questions, he answers no questions right?

C. What is the probability that out of 19 questions, he answers exactly one question right?

Explore the probability of answering multiple-choice questions randomly. Calculate the likelihood of a student answering none or exactly one question correctly out of 19 questions. Understand the concepts of binomial distribution and probability.

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