Q.In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that the student knows the answer given that he answered it correctly?

Concept: Conditional Probability (Bayes' Theorem)

Let $K$ = event that the student knows the answer, $G$ = event that he guesses, and $C$ = event that he answers correctly.

We are given:

• $P(K) = \frac{3}{4}$, $P(G) = \frac{1}{4}$
• $P(C \mid G) = \frac{1}{4}$, and if he knows, he is certainly correct: $P(C \mid K) = 1$

We want $P(K \mid C)$.

Step 1: Find $P(C)$ using the law of total probability:

$P(C) = P(K) \cdot P(C \mid K) + P(G) \cdot P(C \mid G) = \frac{3}{4} \cdot 1 + \frac{1}{4} \cdot \frac{1}{4} = \frac{3}{4} + \frac{1}{16} = \frac{12}{16} + \frac{1}{16} = \frac{13}{16}$ …

The key here is conditional probability: we want $P(\text{knows} \mid \text{correct})$. The event “answered correctly” can happen in two mutually exclusive ways — either the student knows the answer (and is therefore always correct), or they guess (and are correct only with probability $\frac14$).

Bayes’ theorem lets us reverse the conditional: we know $P(\text{correct} \mid \text{knows}) = 1$ and $P(\text{correct} \mid \text{guesses}) = \frac14$, and we know the prior probabilities $\frac34$ and $\frac14$ respectively. The theorem combines these to give the “updated” probability after seeing the correct answer.

1. Define the events clearly Let $K$ = “student knows the answer”, and $C$ = “student answers correctly”. We are given:

2. Find the total probability of answering correctly By the law of total probability:

3. Apply Bayes’ theorem …

Method: Bayes' Theorem when one branch is a certainty

This is the standard "given the observed outcome, find which route produced it" pattern, with one twist that appears often: one of the routes guarantees the outcome.

Steps

Step 1: List the routes and their priors.

Identify the exhaustive ways the situation could have started — "knows" vs "guesses" — with priors $P(H_i)$.

Step 2: Assign each route's likelihood — and spot the certain one.

Write $P(E \mid H_i)$. When a route makes the outcome inevitable (a student who knows is certainly correct), that likelihood is exactly $1$, not some value less than $1$. Getting this branch right is the crux of the problem. …

Common Mistakes

Mistake 1: Using $P(\text{correct}\mid \text{knows}) < 1$.

Why it's wrong: a student who knows the answer is certain to be correct, so this likelihood is exactly $1$. Correct approach: set $P(C\mid K)=1$; only the guessing branch carries $\tfrac14$.

Mistake 2: Dropping the guessing branch from $P(C)$.

Why it's wrong: a correct answer can also come from a lucky guess, so $P(C)$ must include it. Correct approach: $P(C)=\tfrac34\cdot1+\tfrac14\cdot\tfrac14=\tfrac{13}{16}$. …