Math Problem Statement

Suppose that in answering a question on a multiple-choice test an examinee either

knows as answer or he guesses. Let p be the probability that he will know the answer,

and let 1 − p be the probability that he will guess. Assume that the probability

of answering a question correctly is unity for an examinee who knows the answer

the answer and 1/m for an examinee who guesses, where m is the number of M-C

alternatives. Find the conditional probability that an examinee knew the answer to

a question given that he has correctly answered it.

Solution

This problem involves applying Bayes' Theorem to find the conditional probability that an examinee knew the answer given that he has answered the question correctly.

Let's define the events:

  • KK: The event that the examinee knew the answer.
  • GG: The event that the examinee guessed the answer.
  • CC: The event that the examinee correctly answered the question.

Given information:

  • P(K)=pP(K) = p: Probability that the examinee knew the answer.
  • P(G)=1pP(G) = 1 - p: Probability that the examinee guessed the answer.
  • P(CK)=1P(C|K) = 1: Probability of answering correctly given that the examinee knew the answer.
  • P(CG)=1mP(C|G) = \frac{1}{m}: Probability of answering correctly given that the examinee guessed the answer (where mm is the number of multiple-choice alternatives).

Objective:

We want to find the conditional probability P(KC)P(K|C), which is the probability that the examinee knew the answer given that he answered the question correctly.

Applying Bayes' Theorem:

P(KC)=P(CK)P(K)P(C)P(K|C) = \frac{P(C|K) \cdot P(K)}{P(C)}

Where P(C)P(C) is the total probability of answering correctly. We can find P(C)P(C) using the law of total probability:

P(C)=P(CK)P(K)+P(CG)P(G)P(C) = P(C|K) \cdot P(K) + P(C|G) \cdot P(G)

Substituting the given values:

P(C)=(1)p+(1m)(1p)P(C) = (1) \cdot p + \left(\frac{1}{m}\right) \cdot (1 - p)

So:

P(C)=p+1pmP(C) = p + \frac{1 - p}{m}

Now, substituting into Bayes' Theorem:

P(KC)=1pp+1pmP(K|C) = \frac{1 \cdot p}{p + \frac{1 - p}{m}}

This simplifies to:

P(KC)=pp+1pmP(K|C) = \frac{p}{p + \frac{1 - p}{m}}

Final Answer:

P(KC)=pp+1pmP(K|C) = \frac{p}{p + \frac{1 - p}{m}}

This is the conditional probability that an examinee knew the answer to a question given that he has correctly answered it.

If you have any further questions or need more clarification, feel free to ask!

Related Questions:

  1. How would the conditional probability change if mm were large?
  2. What is the probability that an examinee guesses and still answers correctly?
  3. How does this model change if the probabilities for guessing correctly are not uniform across the options?
  4. How can this result be interpreted in the context of test reliability?
  5. What is the impact of increasing pp on the conditional probability P(KC)P(K|C)?

Tip:

When working with conditional probabilities and Bayes' Theorem, it's crucial to clearly define each event and identify the total probability, which often requires considering all possible scenarios leading to the observed outcome.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Bayes' Theorem

Formulas

Bayes' Theorem

Theorems

Bayes' Theorem

Suitable Grade Level

Advanced High School