Q.Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Concept: Conditional Probability — we need $P(\text{Hostel} \mid \text{A grade})$, found using Bayes' theorem.

Step 1 — Define events and given probabilities

Let $H$ = student is a hosteller, $D$ = day scholar, $A$ = attains A grade.

$P(H) = 0.6$, $P(D) = 0.4$, $P(A \mid H) = 0.3$, $P(A \mid D) = 0.2$.

Step 2 — Find total probability of A grade

$P(A) = P(H)P(A \mid H) + P(D)P(A \mid D) = (0.6)(0.3) + (0.4)(0.2) = 0.18 + 0.08 = 0.26$.

Step 3 — Apply Bayes' theorem

$P(H \mid A) = \frac{P(H)P(A \mid H)}{P(A)} = \frac{0.18}{0.26} = \frac{18}{26} = \frac{9}{13}$.

Why Bayes’ theorem?

We are told two things about the college:

• 60% of students are hostellers, 40% are day scholars.
• Among hostellers, 30% get A grade; among day scholars, 20% get A grade.

But the question flips the direction: given that a randomly chosen student has an A grade, what is the probability they are a hosteller? That is a classic inverse probability problem — we know $P(\text{A} \mid \text{Hostel})$ and want $P(\text{Hostel} \mid \text{A})$.

Bayes’ theorem is the tool for exactly this: it lets us “reverse” the condition using the overall probabilities.

The denominator $P(\text{A})$ is the total probability of getting an A grade, which we find by the law of total probability — summing over the two groups.

Step-by-step solution

1. Define events clearly

Let $H$ = student is a hosteller, $D$ = student is a day scholar, and $A$ = student gets A grade.

From the problem:

• $P(H) = 0.6$, $P(D) = 0.4$
• $P(A \mid H) = 0.3$, $P(A \mid D) = 0.2$

2. Find the total probability of A grade

A student can get an A either as a hosteller or as a day scholar. These are mutually exclusive and cover all students, so:

Substitute:

So 26% of all students get an A grade.

3. Apply Bayes’ theorem

We want $P(H \mid A)$:

Simplify the fraction:

4. Interpret the result

Even though hostellers are a majority (60%), their A-grade rate (30%) is only moderately higher than day scholars’ (20%). So when we see an A-grade student, the chance they are a hosteller is $\frac{9}{13} \approx 0.6923$, or about 69.2%.

Method: Bayes' Theorem for a "reverse the direction" question

Use this when a problem hands you probabilities in one direction — group membership → outcome rate — but asks the opposite direction, outcome → which group.

Steps

Step 1: Split the population into its groups and record the priors.

Identify the exhaustive categories $H_i$ (hosteller / day scholar) and their shares $P(H_i)$.

Step 2: Record the conditional rate of the observed trait in each group.

Write $P(E \mid H_i)$ — the rate of the observed event (A grade) inside each group. These are the "forward" numbers the problem gives directly.

Step 3: Build the overall rate of the trait (law of total probability).

a weighted average of the group rates, weighted by group size.

Step 4: Invert with Bayes' theorem.

Because the larger group's prior sits in the numerator, the answer usually differs from a naive comparison of the raw group rates — the priors genuinely matter.

Common Mistakes

Mistake 1: Ignoring the base rates and comparing only the A-grade rates.

Why it's wrong: taking $\dfrac{0.3}{0.3+0.2}=0.6$ throws away the $60\%$/$40\%$ group sizes. Correct approach: weight each rate by its group's prior before dividing — that is what Bayes' theorem does, giving $\tfrac{9}{13}$.

Mistake 2: Leaving the day-scholar term out of $P(A)$.

Why it's wrong: an A grade can come from either group, so the total must include both. Correct approach: $P(A)=(0.6)(0.3)+(0.4)(0.2)=0.26$.

Mistake 3: Inverting the wrong conditional.

Why it's wrong: $P(H\mid A)$ is asked, not $P(A\mid H)=0.3$. Correct approach: keep the target direction "given A grade, find hosteller" and apply Bayes' theorem.