Which theorem is used to find P(bus | late)?

Bayes' Theorem: P(H|E) = P(H)·P(E|H) / P(E), where P(E) is found using the Total Probability Theorem.

What is P(Late) using Total Probability Theorem?

P(L) = P(B)·P(L|B) + P(S)·P(L|S) + P(C)·P(L|C) = (2/5)(1/5)+(1/5)(1/3)+(2/5)(1/4) = 2/25+1/15+1/10 = 12/150+10/150+15/150 = 37/150.

What is P(Bus | Late)?

P(B|L) = P(B)·P(L|B)/P(L) = (2/25)/(37/150) = (2/25)×(150/37) = 300/925 = 12/37.

How to verify the answer?

P(B|L)+P(S|L)+P(C|L) must equal 1. P(S|L)=(1/15)/(37/150)=10/37, P(C|L)=(1/10)/(37/150)=15/37. Sum=12/37+10/37+15/37=37/37=1 ✓

Candidate travels by bus 2/5, scooter 1/5, car 2/5. Late probabilities 1/5, 1/3, 1/4. Given late, probability of bus

JEERANKERS

MathProbability

Q MCQ Probability

A candidate travels by bus, scooter, or car with probabilities $\dfrac{2}{5},\dfrac{1}{5},\dfrac{2}{5}$ respectively. The probabilities of reaching late are $\dfrac{1}{5},\dfrac{1}{3},\dfrac{1}{4}$ if he uses bus, scooter, car respectively. Given that the candidate reached late, the probability that he travelled by bus is:

A) $\dfrac{11}{37}$ B) $\dfrac{12}{37}$ C) $\dfrac{13}{37}$ D) $\dfrac{14}{37}$

✅ Correct Answer

B) 12/37

✓

Solution

Define all events and given probabilities

$P(B)=\dfrac{2}{5},\quad P(S)=\dfrac{1}{5},\quad P(C)=\dfrac{2}{5}$

$P(L|B)=\dfrac{1}{5},\quad P(L|S)=\dfrac{1}{3},\quad P(L|C)=\dfrac{1}{4}$

We need $P(B|L)$ — probability of bus given late.

Find P(L) using Total Probability Theorem

$P(L)=P(B)\cdot P(L|B)+P(S)\cdot P(L|S)+P(C)\cdot P(L|C)$

$=\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{3}+\dfrac{2}{5}\cdot\dfrac{1}{4}$

$=\dfrac{2}{25}+\dfrac{1}{15}+\dfrac{1}{10}$

LCM of 25, 15, 10 is 150:

$=\dfrac{12}{150}+\dfrac{10}{150}+\dfrac{15}{150}=\dfrac{37}{150}$

Apply Bayes' Theorem

$$P(B|L)=\frac{P(B)\cdot P(L|B)}{P(L)}=\frac{\dfrac{2}{5}\cdot\dfrac{1}{5}}{\dfrac{37}{150}}=\frac{\dfrac{2}{25}}{\dfrac{37}{150}}=\frac{2}{25}\times\frac{150}{37}=\frac{300}{925}=\boxed{\dfrac{12}{37}}$$

Verify: all posterior probabilities sum to 1

$P(S|L)=\dfrac{1/15}{37/150}=\dfrac{10}{37}$

$P(C|L)=\dfrac{1/10}{37/150}=\dfrac{15}{37}$

$P(B|L)+P(S|L)+P(C|L)=\dfrac{12}{37}+\dfrac{10}{37}+\dfrac{15}{37}=\dfrac{37}{37}=1\ \checkmark$

📘 Key Concept — Bayes' Theorem

$$P(H_i|E)=\frac{P(H_i)\cdot P(E|H_i)}{\sum_{j}P(H_j)\cdot P(E|H_j)}$$ The denominator is $P(E)$ computed by the Total Probability Theorem. This is one of the most frequently tested topics in JEE Main probability.

💡

Important Concepts

Prior vs Posterior Probability $P(B)=2/5$ is the prior (before any observation). $P(B|L)=12/37$ is the posterior (after observing "reached late"). Bayes converts priors to posteriors using evidence.

Total Probability Theorem If $H_1,H_2,H_3$ are mutually exclusive and exhaustive: $P(E)=\sum P(H_i)\cdot P(E|H_i)$. Always verify that the hypotheses are exhaustive: $P(B)+P(S)+P(C)=2/5+1/5+2/5=1$ ✓

LCM Trick for Fractions To add $\dfrac{2}{25}+\dfrac{1}{15}+\dfrac{1}{10}$, find LCM(25,15,10)=150. Convert: $\dfrac{12}{150}+\dfrac{10}{150}+\dfrac{15}{150}=\dfrac{37}{150}$. Keeping a common denominator avoids decimal errors.

Quick Verification After finding all three posterior probabilities, they must sum to 1. Here $12+10+15=37$, so $12/37+10/37+15/37=1$ ✓. This is a reliable check in exam conditions.

FAQs

What is Bayes' Theorem intuitively?

⌄

It updates our belief about a hypothesis when new evidence arrives. We knew the candidate takes bus with probability 2/5. After learning he's late (evidence), we update this belief to 12/37 ≈ 32%, slightly less than the prior 40%, because the bus has a lower late-probability than the car.

Why is the bus posterior (12/37) less than the prior (2/5)?

⌄

Because $P(L|B)=1/5$ is the lowest among all modes. Being late is less likely if you take bus, so observing "late" is evidence against having taken the bus, reducing its posterior probability.

What is P(scooter | late)?

⌄

$P(S|L)=\dfrac{P(S)\cdot P(L|S)}{P(L)}=\dfrac{(1/5)(1/3)}{37/150}=\dfrac{1/15}{37/150}=\dfrac{10}{37}$.

Can I use a tree diagram for this problem?

⌄

Yes! Draw three branches (Bus, Scooter, Car) with their probabilities, then sub-branches (Late, Not Late) with conditional probabilities. Bayes' answer = (branch with Bus→Late) ÷ (total Late branches) = (2/5 × 1/5) ÷ (37/150) = 12/37.

Is this from JEE Main 2026?

⌄

Yes, this question appeared in JEE Main 2026.

🔗