Part (a): Probability of Securing a Distinction

Let’s break down the probability of the student getting a distinction.

1. Total Questions: There are 8 questions.

2. Answers per Question: Each question has 3 possible answers, with only one being correct.

3. Student’s Selection Method: The student rolls a die to pick an answer, and each answer has a $\frac{1}{3}$ probability of being selected. Thus, each answer the student picks has a probability of $\frac{1}{3}$ of being correct.

4. Defining a Distinction: To achieve a distinction, the student must get at least 75% of the answers correct. This means the student needs to answer at least: $8 \times 0.75 = 6 \text{ questions correctly.}$

5. Defining Success Probability for Each Question: Each answer has a $\frac{1}{3}$ probability of being correct.

6. Modeling as a Binomial Distribution: Let $X$ be the number of correct answers. Then $X$ follows a binomial distribution with parameters: $X \sim \text{Binomial}(n = 8, p = \frac{1}{3})$ We want $P(X \geq 6)$.

7. Calculating $P(X \geq 6)$: The probability of securing a distinction is: $P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8)$ We can compute each term using the binomial probability formula: $P(X = k) = \binom{8}{k} \left( \frac{1}{3} \right)^k \left( \frac{2}{3} \right)^{8-k}$

   Calculating these values can give us the probability of achieving a distinction.

Part (b): Probability of No Errors in 5 Pages

Let’s solve this using the Poisson distribution.

1. Total Pages and Errors: In a 520-page book, there are 390 typographical errors.

2. Average Errors per Page (λ): The average number of errors per page can be found by dividing the total errors by the total pages: $\lambda = \frac{390}{520} = 0.75$

3. Using Poisson Distribution: Let $Y$ be the number of errors in a random sample of 5 pages. The Poisson distribution is suitable here, with a mean number of errors over 5 pages being: $\lambda_{\text{5 pages}} = 5 \times 0.75 = 3.75$ Thus, $Y \sim \text{Poisson}(3.75)$.

4. Finding Probability of No Errors in 5 Pages: We need $P(Y = 0)$: $P(Y = 0) = \frac{e^{-\lambda_{\text{5 pages}}} \lambda_{\text{5 pages}}^0}{0!} = e^{-3.75}$ Calculating $e^{-3.75}$ will give us the probability that a random sample of 5 pages contains no errors.

Would you like to see these calculations in detail, or do you have any questions?

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Additional Questions to Explore

1. How would the probability change in part (a) if there were 4 answers instead of 3?
2. What is the probability of getting exactly 5 correct answers in part (a)?
3. How would the outcome change if the student needed only a 50% score to pass?
4. How does the mean number of errors per page affect the Poisson probability in part (b)?
5. If we consider a larger sample of pages in part (b), how would the probability of no errors change?

Tip

For Poisson problems, keep in mind that the mean number of events within the given interval is key for calculating probabilities directly using the Poisson formula.