# In a test, an examinee either guesses or copies or knows that answer to a multiple-choice question which has four choices. The probability that he makes a guess is 1/3, and the probability that he copies is 1/6. The probability that his answer is correct, given the copied it, is 1/8. Find the probability that he knew the answer to the question, given that he answered it correctly.

Probability is a concept used in math and science to know the likelihood or occurrence of an event.

## Answer: The probability that he knew the answer to the question, given that he answered correctly is 24/29.

Let's find the probability that he knew the answer to the question, given that he answered it correctly.

**Explanation:**

Let us understand the scenario better by defining the following events:

A: The examinee guess the answer

⇒ P(A) = 1/3

B: The examinee copies the answer

⇒ P(B) = 1/6

C: The examinee knows the answer

⇒ P(C) = 1 - (1/3 + 1/6) = 1 - 1/2 = 1/2 {as the sum of probabilities = 1}

S: The answer is correct

Let us understand the given conditional probabilites also:

P(S | A): Probability that examinee got the answer correct by guessing = 1/4 {as the question has 4 choices}

P(S | B): Probability that examinee got the answer correct by copying = 1/8

P(S | C): Probability that examinee got the answer correct by knowing the answer = 1 {as it is a sure event}

P(C | S): Probability that examinee knows the answer given that he answered correctly

Therefore by Bayes' theorem:

\begin{equation}

\text{P}\left(\frac{\text{C}}{\text{S}}\right)=\frac{\text{P}(\frac{\text{S}}{\text{C}}) \cdot \text{P}(\text{C})}{\text{P}(\frac{\text{S}}{\text{A}}) \cdot \text{P}(\text{A}) + \text{P}(\frac{\text{S}}{\text{B}}) \cdot \text{P}(\text{B})+ \text{P}(\frac{\text{S}}{\text{C}}) \cdot \text{P}(\text{C})}=\frac{1\cdot \frac{1}{2}}{\frac{1}{4}\cdot\frac{1}{3} + \frac{1}{8}\cdot\frac{1}{6} + \frac{1}{2}\cdot 1} = \frac{24}{29}

\end{equation}