homework - Bayes' theorem problem In a test, an examinee either guesses or copies of knows the answer to a multiple choice question with four choices. The probability that he makes the guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct, given that he copied it is $\frac{1}{8}$. Find the probability that he knew the answer to the question, given that he correctly answered it.
My Approach:
Let $G$ be the event of guessing an answer, $C$ be the event of copying an answer, $A$ be the event of knowing the answer and $T$ be the event of answering correctly. 
Given, $$P(G)=\frac{1}{3};P(C)=\frac{1}{6};P(T|C)=\frac{1}{8}$$
$$\therefore P(A)=1-(P(G)+P(C))=\frac{1}{2}$$
By Bayes' theorem,
$$P(A|T)=\frac{P(T|A)\times P(A)}{P(T|A)\times P(A) + P(T|G)\times P(G)+P(T|C)\times P(C)}$$

I'm not able to approach further after this step. This is because I don't know that value of $P(T|A)$ and $P(T|G)$. What am I missing exactly? Or there's flaw in the question?
 A: It is important to define the events of interest clearly and unambiguously.
Using your existing notation, there three actions:
$$C = \text{copied an answer choice} \\
A = \text{knows (and chooses) the correct answer} \\
G = \text{guesses an answer choice} \\
$$
and two outcomes of those actions:
$$T = \text{the answer chosen is correct} \\
\bar T = \text{the answer chosen is incorrect} \\
$$
We are given the following probabilities:
$$\Pr[C] = \frac{1}{6}; \quad \Pr[G] = \frac{1}{3}; \quad \Pr[A] = 1 - \Pr[C] - \Pr[G] = \frac{1}{2}; \\
\Pr[T \mid C] = \frac{1}{8}.$$  There are also two additional probabilities implied by the statement of the problem:
$$\Pr[T \mid G] = \frac{1}{4}; \quad \Pr[T \mid A] = 1.$$
This is because, if the test-taker guesses, and there are four answer choices, assuming that the guess is completely at random, the chance of having guessed correctly is $1$ out of $4$.  And of course, if the test-taker knows the correct answer, the probability of choosing the correct answer is $1$.
The desired probability is $$\Pr[A \mid T] = \frac{\Pr[T \mid A] \Pr[A]}{\Pr[T]}$$ by Bayes' theorem.  The only probability on the RHS we do not already have is $\Pr[T]$, the unconditional probability of a correct answer.  This is found through the law of total probability:  $$\Pr[T] = \Pr[T \mid C]\Pr[C] + \Pr[T \mid A]\Pr[A] + \Pr[T \mid G]\Pr[G],$$ as we condition on the possible set of actions (for which there are only those three, and each action is mutually exclusive).  All of these probabilities are known, so the rest is simply arithmetic.
A: 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}$
