A multiple choice exam has 10 questions. Each question has 3 possible answers, of which one is correct. A student knows the correct answers to 4 questions and guesses the answers to the other 6 questions.

It turns out that the student answered the first question correctly. What are the chances that the student was merely guessing?

1 point

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(13)(410)+(1)(610)(13)(410)(31​)(104​)(31​)(104​)+(1)(106​)​

(13)(410)(13)(410)+(1)(610)(31​)(104​)+(1)(106​)(31​)(104​)​

(13)(610)(13)(610)+(1)(410)(31​)(106​)+(1)(104​)(31​)(106​)​

To solve this problem, we need to use Bayes' theorem to find the probability that the student was guessing given that they answered the first question correctly.

Let's denote:
- \( G \) as the event that the student was guessing on the first question.
- \( K \) as the event that the student knew the answer to the first question.
- \( C \) as the event that the student answered the first question correctly.

We need to find \( P(G|C) \), the probability that the student was guessing given that they answered the first question correctly.

By Bayes' theorem:
\[ P(G|C) = \frac{P(C|G) \cdot P(G)}{P(C)} \]

First, let's calculate each component:

1. **\( P(G) \):** The probability that the student was guessing on the first question. Since the student knows the answers to 4 out of 10 questions and guesses on the remaining 6:
\[ P(G) = \frac{6}{10} = \frac{3}{5} \]

2. **\( P(K) \):** The probability that the student knew the answer to the first question.
\[ P(K) = \frac{4}{10} = \frac{2}{5} \]

3. **\( P(C|G) \):** The probability of answering correctly given that the student was guessing. Since there is only one correct answer out of three possible answers:
\[ P(C|G) = \frac{1}{3} \]

4. **\( P(C|K) \):** The probability of answering correctly given that the student knew the answer.
\[ P(C|K) = 1 \]

5. **\( P(C) \):** The total probability of answering the first question correctly. This is calculated by considering both possibilities (knowing the answer and guessing):
\[ P(C) = P(C|G)P(G) + P(C|K)P(K) \]
\[ P(C) = \frac{1}{3} \cdot \frac{3}{5} + 1 \cdot \frac{2}{5} \]
\[ P(C) = \frac{1}{5} + \frac{2}{5} \]
\[ P(C) = \frac{3}{5} \]

Now, applying Bayes' theorem:
\[ P(G|C) = \frac{P(C|G) \cdot P(G)}{P(C)} = \frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{3}{5}} = \frac{\frac{1}{3}}{1} = \frac{1}{3} \]

Thus, the probability that the student was merely guessing given that they answered the first question correctly is \( \frac{1}{3} \).

Do you want more details or have any questions?

Here are 5 more questions you might find interesting:
1. How do you apply Bayes' theorem to different types of probability problems?
2. Can you explain the concept of prior and posterior probabilities?
3. How does conditional probability differ from joint probability?
4. What are some common real-life applications of Bayes' theorem?
5. How can you calculate the probability of dependent events in a sequence?

**Tip:** When dealing with probabilities, always carefully consider all possible scenarios and how they contribute to the total probability.

Understand how to apply Bayes' theorem to calculate the probability that a student was guessing on a multiple choice exam. This problem involves 10 questions with 3 possible answers each, where the student knew 4 answers and guessed on 6. Learn the step-by-step solution and key mathematical concepts involved. Suitable for advanced high school students.

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