To solve the problem of determining the probability that a randomly chosen student who answered both questions correctly was poorly prepared, we need to use conditional probability principles. Here's how this can be approached step by step:

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1. Understand the Problem

• Define the events:
  • $A$: Student answered both questions correctly.
  • $B$: Student is poorly prepared.
• We are looking for $P(B \mid A)$, the probability a student was poorly prepared given they answered both questions correctly.

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2. Apply Bayes' Theorem

Bayes' theorem is expressed as: $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$

Where:

• $P(A \cap B)$: Probability a poorly prepared student answered both questions correctly.
• $P(A)$: Total probability a student answered both questions correctly, considering all preparation levels.

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3. Break Down $P(A)$

The total probability $P(A)$ is calculated using the law of total probability: $P(A) = P(A \mid B)P(B) + P(A \mid B^c)P(B^c)$ Where:

• $P(A \mid B)$: Probability a poorly prepared student answered both questions correctly.
• $P(B)$: Probability a student is poorly prepared.
• $P(A \mid B^c)$: Probability a well-prepared student answered both questions correctly.
• $P(B^c)$: Probability a student is well-prepared ($1 - P(B)$).

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4. Substitute Probabilities

• Insert specific probabilities into the formulas based on the given data. For example:
  • If poorly prepared students answer both questions correctly 30% of the time: $P(A \mid B) = 0.3$.
  • If well-prepared students answer both correctly 90% of the time: $P(A \mid B^c) = 0.9$.
  • If 40% of students are poorly prepared: $P(B) = 0.4$, so $P(B^c) = 0.6$.

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5. Perform the Calculations

Calculate $P(A)$: $P(A) = (0.3)(0.4) + (0.9)(0.6)$ $P(A) = 0.12 + 0.54 = 0.66$

Calculate $P(A \cap B)$: $P(A \cap B) = P(A \mid B)P(B) = (0.3)(0.4) = 0.12$

Calculate $P(B \mid A)$: $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.66} \approx 0.182$

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Conclusion

The probability that a randomly chosen student who answered both questions correctly was poorly prepared is approximately 18.2%.

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Would you like further details, or do you have specific questions about this process?

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Related Questions:

1. How would the probability change if the preparation levels were different?
2. What does this probability tell us about the students' overall performance?
3. How does the assumption about $P(A \mid B)$ affect the final probability?
4. Can we verify this calculation using simulation or a spreadsheet?
5. What are the implications of a high $P(B \mid A)$ in real-world settings?

Tip:

Always double-check the consistency of given probabilities to ensure calculations align logically!