Q.$A$ and $B$ are two students. Their chances of solving a problem correctly are $\dfrac{1}{3}$ and $\dfrac{1}{4}$, respectively. If the probability of their making a common error is $\dfrac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is
(A) $\dfrac{1}{12}$
(B) $\dfrac{1}{40}$
(C) $\dfrac{13}{120}$
(D) $\dfrac{10}{13}$

Concept: Conditional Probability — we want $P(\text{correct} \mid \text{same answer})$.

Step 1: Define events.

Let $C$ = both answer correctly, $W$ = both answer wrongly (independently), and $E$ = they make the same error (given as $\frac{1}{20}$).

$P(C) = \frac13 \cdot \frac14 = \frac{1}{12}$.

$P(W) = \frac23 \cdot \frac34 = \frac{6}{12} = \frac12$.

Step 2: Probability they get the same answer.

Same answer happens if: both correct (no error), OR both wrong and they make the same error.

$P(\text{same answer}) = P(C) + P(W) \cdot P(E) = \frac{1}{12} + \frac12 \cdot \frac{1}{20} = \frac{1}{12} + \frac{1}{40} = \frac{10}{120} + \frac{3}{120} = \frac{13}{120}$. …

We want $P(\text{answer correct}\mid \text{they give the same answer})$. Two independent students can arrive at the same answer in two mutually exclusive ways: both solve it correctly, or both make the same wrong answer.

Given: $P(A\text{ correct})=\dfrac{1}{3}$, $P(B\text{ correct})=\dfrac{1}{4}$, and the chance that (when both are wrong) their errors coincide is $\dfrac{1}{20}$.

Step 1 — Both correct. By independence,

Step 2 — Both wrong with the same error.

and the common error occurs with probability $\dfrac{1}{20}$, so

Step 3 — Probability of the same answer. These two cases are disjoint: …

Method: Bayes' theorem via total probability for a "same result" question

Use this when an outcome can arise through two or more disjoint routes and you must find the probability that it came from a particular route, given the outcome happened ("both students got the same answer — is it correct?").

Steps

Step 1: List the disjoint routes to the observed event.

"Same answer" happens either because both are correct, or because both are wrong with the same error. These cases cannot overlap.

Step 2: Find the probability of each route (multiplication theorem).

Use independence within each route: $P(\text{both correct}) = P(A)\,P(B)$, and $P(\text{both wrong}) = \big(1-P(A)\big)\big(1-P(B)\big)$, then multiply by the given chance of a common error.

Step 3: Total probability of the observed event. …

Common Mistakes

Mistake 1: Forgetting the "both wrong with the same error" route.

Why it's wrong: if the denominator is taken as just $P(\text{both correct})=\tfrac1{12}$, the conditional probability comes out as $1$. Correct approach: the same answer can also arise from a shared error, so $P(\text{same answer})=\tfrac1{12}+\tfrac1{40}=\tfrac{13}{120}$.

Mistake 2: Reporting $P(\text{both correct})=\tfrac1{12}$ as the final answer.

Why it's wrong: the question asks for a conditional probability given the same answer, not the joint probability of both being correct. Correct approach: divide by the total: $\dfrac{1/12}{13/120}=\dfrac{10}{13}$. …