Probability using Bayes rule

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Suppose that in answering a question on a true/false test, an examinee either knows the answer with probability $p$ or s/he guesses with probability $1-p$. Assume that if the examinee knows the answer to a question, the probability that s/he gives the correct answer is $1$, and if s/he guesses then s/he only gives the correct answer with probability $0.5$.

Use Bayes rule to compute the probability that an examinee knew the answer to a question given that s/he has correctly answered it.

First I wrote out all the probabilities from the question.

$$ P(\text{Wrong}) = 0.5 \\ P(\text{Correct}) = 0.5 \\ P(\text{Correct} \mid \text{Known}) = 1 \\ P(\text{Wrong} \mid \text{Known}) = 0 \\ P(\text{Correct} \mid \text{Guess}) = 0.5 \\ P(\text{Wrong} \mid \text{Guess}) = 0.5 \\ $$ I tried creating two equations with two unknowns to get a value of $p$ shown below:

$$ (1)\quad 0.5 = \frac{P(\text{Guess} \mid \text{Correct})\cdot0.5}{1 - p} $$

$$ (2)\quad 1 = \frac{P(\text{Known} \mid \text{Correct})\cdot0.5}{p} $$

Then rearranged $(2)$ to get the following:

$$ P(\text{Known} \mid \text{Correct}) = 2p $$

And $P(\text{Guesses} \mid \text{Correct})$ is equal to $1 - P(\text{Knows} \mid \text{Correct})$ so I substituted that back into $(1)$

$$ (1)\quad 0.5 = \frac{(1 - 2p)\cdot0.5}{1 - p}\\ (1)\quad 0.5 = \frac{0.5 - p}{1 - p}\\ (1)\quad 0.5 - 0.5p = 0.5 - p\\ 0.5p = 0\\ p = 0 $$

But this is can't be right, the question is only 5% so doesn't seem like it would be this much work, am I missing something simple here? The main equation that needs to be solved:

$$ P(\text{Known} \mid \text{Correct}) = \frac{P(\text{Correct} \mid \text{Known}) P(\text{Known})}{P(\text{Correct})} $$

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Let's write the question out:

$$ \mathcal{P}(\text{Knows the answer}) = p \qquad \mathcal{P}(\text{Doesn't know the answer}) = 1-p$$

And

$$ \mathcal{P}(\text{Correct} | \text{Knows the answer}) = 1; \;\mathcal{P}(\text{Wrong} | \text{Knows the answer}) = 0$$

$$ \mathcal{P}(\text{Correct} | \text{Doesn't know the answer}) = 0.5; \;\mathcal{P}(\text{Wrong} | \text{Doesn't know the answer}) = 0.5$$

\par The question is to find: $$\mathcal{P}(\text{Knows the answer} | \text{Correct}) = ? $$

Using Bayes, one finds: $$\begin{align}\mathcal{P}(\text{Knows the answer} | \text{Correct}) &= \dfrac{\mathcal{P}(\text{Correct} | \text{Knows the answer})\mathcal{P}(\text{Knows the answer)} }{\mathcal{P}(\text{Correct})} \\ &= \dfrac{1\cdot p}{\mathcal{P}(\text{Correct})} \end{align}$$

Now using the law of total probability, $$\begin{align}\mathcal{P}(\text{Correct}) &= \mathcal{P}(\text{Correct} | \text{Knows the answer})\cdot \mathcal{P}(\text{Knows the answer}) \\ &+\mathcal{P}(\text{Correct} | \text{Doesn't know the answer})\cdot \mathcal{P}(\text{Doesn't know the answer})\\ & = 1\cdot p+0.5\cdot (1-p) \\ & = 0.5p +0.5 \end{align}$$

Which implies: $$\begin{align}\mathcal{P}(\text{Knows the answer} | \text{Correct}) &= \dfrac{p}{0.5p+0.5} \end{align}$$

Notice how when $p=1$ (which implies the examinee knows the answer all the time) then this probability also equals 1. (which makes sense)

To visualize this probability in function of $p$, see the following graph: Probability

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Question 1:

In answering a question on a multiple-choice test, an examinee either knows the answer (with probability p), or he Guesses (with probability 1 - p). Assume that the probability of answering a question correctly is unity for an examinee who knows the answer and 1/m for the examinee who guesses, where m is the number of multiple-choice alternatives. Supposing an examinee answers a question correctly, what is the probability that he really knows the answer?

Solution :

MCQ : m options.

P(KNOWS the correct answer) : p
P(GUESSES the correct answer) : (1 - p)

The probability of answering a question correctly is unity for an examinee who knows the answer.
A = The examinee answers CORRECTLY.
Let K = The examinee KNOWS the answer.
Then , $P(\frac{A}{K}) = 1$

The probability of answering a question correctly is 1/m for the examinee who GUESSES, where m is the number of multiple-choice alternatives.
A = The examinee answers correctly.
Let G = The examinee GUESSES the answer.
Then, $P(\frac{A}{G}) = \frac{1}{m}$

Then, the conditional probability that a man knew the answer to a question, given that he has Correctly answered it, is equal to $P (K | A ) = P( \frac{\text{Man knew the answer to the Question}}{\text{He has correctly answered it}}) = P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question OR He Guessed the answer }} )= P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question + He Guessed the answer }} ) =\frac{p(1)}{p(1) + (1-p)\frac{1}{m}} = \frac{mp}{mp + 1- p}$

Now If we add 1 more condition of Copying. Then, Let us look at this Question
Question 2: In a test, an examinee, either Guesses Or Copies Or Knows the answer for multiple-choice test having 4 options of which only 1 is correct.The probability that he makes a guess is 1/3 and the probability for copying is 1/6. The probability that his answer is correct given that he copied it is 1/8. Prove that The probability that he knew the answer, given that his answer is correct is 24/29.

Solution :

Let, C be the probability that he will COPY the answer.
C = $\frac{1}{6}$
A = The examinee answers CORRECTLY.
Then, $P(Correct|Copy) = P(A|C) =(\frac{1}{8})$ The probability of answering a question correctly is 1/m for the examinee who GUESSES, where m is the number of multiple-choice alternatives.
A = The examinee answers correctly.
Let G = The examinee GUESSES the answer. = 1/3
Then, $P(\frac{A}{G}) = \frac{1}{m} = \frac{1}{4}$

Let K = The examinee KNOWS the answer.
Then $K = 1 - (G+C) = 1 - (\frac{1}{6} + \frac{1}{3}) = \frac{1}{2}$

Here also, we will say: the Probability that his answer is correct given that he KNOWS the answer => $P(A|K) = 1 $.
The probability that he knew the answer, given that his answer is correct =

$ P( \frac{\text{Man knew the answer to the Question}}{\text{He has correctly answered it}}) = P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question OR He Guessed the answer OR He Copied the correct answer}} )= P(\frac{\text{Man knew the answer to the Question}}{\text{Man knew the answer to the Question + He Guessed the answer + He Copied the correct answer}} ) => P(K|A) = \frac{P(K).P(A|K)}{P(K).P(A|K) + P(G).P(A|G) + P(C).P(A|C)} = \frac{P(K).(1)}{P(K).(1) + P(G).(\frac{1}{options}) + P(C).(\frac{1}{8})} = \frac{\frac{1}{2}.(1)}{\frac{1}{2}.(1) + \frac{1}{3}.(\frac{1}{4}) + \frac{1}{6}.(\frac{1}{8})} = \frac{24}{29}$