P(Train | Reached Late) by Bayes Theorem – Probability JEE Main 2026 PYQ

Nishant Kumar Gupta

Former Faculty · Allen · Aakash · Narayana

Maths mentor behind PadhoLikhoJEE and Quantum Chemistry Classes (Arrah, Bihar). Teaches the “why” behind every JEE trap — not just the trick.

Quick Summary

Question Type:

Conditional probability (Bayes reversal)

Chapter:

Probability

Difficulty:

⭐⭐⭐ Medium

Time to Solve:

90 seconds

Key Concept:

Total probability over the four travel modes gives P(Late) = 18/120 = 3/20; Bayes then gives P(Train | Late) = (9/120)/(18/120) = 1/2

Correct Answer:

Option (1) – 1/2

Why:

Train’s contribution to lateness (9/120) is exactly half of the total lateness probability (18/120)

The Question

JEE Main 2026 (8 April – Evening Shift) – Probability

A candidate has to go to the examination center. The probabilities of travelling by
train, bus, car or other means are 3/10, 1/5, 1/10
and 2/5 respectively. The probabilities of reaching late by these means are
1/4, 1/3, 1/12 and 0 respectively. If the candidate reaches late, the
probability that the candidate travelled by train is:

1. (1) 1/2 ✓
2. (2) 4/9
3. (3) 1/18
4. (4) 2/3

Quick Answer

Correct Option: (1) – 1/2

Reasoning: put all four “late” contributions over the common denominator 120 and Bayes becomes bookkeeping:

• Train: (3/10)(1/4) = 3/40 = 9/120 ··· Bus: (1/5)(1/3) = 1/15 = 8/120
• Car: (1/10)(1/12) = 1/120 ··· Other: (2/5)(0) = 0
• P(Late) = (9 + 8 + 1)/120 = 18/120 = 3/20
• P(Train | Late) = (9/120) / (18/120) = 9/18 = 1/2 ✓ — and the distractors 4/9 and 1/18 are P(Bus | Late) and P(Car | Late), built from the same fractions

Understanding the Concept

The Reversal Structure of Bayes Questions

The question supplies cause-to-effect probabilities — the chance of lateness given each
travel mode — and then asks for the reverse: the probability of a cause (train) given the
effect (late). That reversal is precisely what Bayes’ theorem performs. The denominator is the
law of total probability: late arrivals can only happen through one of the four mutually
exclusive travel modes, so their joint probabilities add up to P(Late). The numerator isolates
the single branch of interest — travelled by train AND reached late.

The Denominator-120 Bookkeeping Trick

Four multiplications with awkward denominators invite arithmetic errors, but every product here
lands on a divisor of 120: train contributes 9/120, bus 8/120, car 1/120, other 0. Writing all
branches over one common denominator converts Bayes into counting units: the late-population
has 18 “units”, of which 9 came by train — the answer 9/18 = 1/2 is visible at a glance. This
unit-bookkeeping presentation is faster and safer than reducing each fraction separately, and
it makes the distractor options transparent: P(Bus | Late) = 8/18 = 4/9 and P(Car | Late) =
1/18 are exactly the neighbouring units. The examiner built wrong options from the right
method applied to the wrong branch — reading the options therefore doubles as a self-check.

The Zero Branch Is Information Too

Travelling by “other means” carries late-probability 0. Students sometimes force it into the
sum anyway and wonder where it fits. Its correct role is to disappear — the branch contributes
nothing to P(Late) — but it still matters structurally: the four modes form a partition whose
probabilities sum to 1 (3/10 + 1/5 + 1/10 + 2/5 = 1), so the three live branches plus the dead
one cover every possibility. A partition check (does the cause list exhaust everything, and do
its probabilities add to 1?) is the 5-second validation that the Bayes setup is legitimate.

The Tree Picture

Draw a first layer with four branches (train, bus, car, other) labelled 3/10, 1/5, 1/10, 2/5,
and a second layer from each with “late” and “on time”. Multiplying along the root-to-late
paths gives the four joint probabilities; P(Late) is their sum; and the conditional answer is
the train path’s share of that sum. Every JEE Bayes question — disease tests, defective
machines, witness reliability — is this same tree with new labels, which is why mastering the
bookkeeping once pays across the entire chapter.

Detailed Step-by-Step Solution

Final Answer

Essential Facts for This Topic

Bayes Toolkit

1. Total probability: for a partition {B₁, …, Bₙ}: P(A) = Σ P(Bᵢ)P(A | Bᵢ)
2. Bayes reversal: P(Bᵢ | A) = P(Bᵢ)P(A | Bᵢ) / Σⱼ P(Bⱼ)P(A | Bⱼ) — numerator is one branch, denominator is the branch sum
3. Partition validity check: the cause probabilities must sum to 1 (here 3/10 + 1/5 + 1/10 + 2/5 = 1)
4. Common-denominator bookkeeping: put every joint term over one LCM (here 120) — the answer becomes unit-counting
5. Zero branches vanish from P(A) but still belong in the partition
6. Option forensics: wrong options in Bayes MCQs are usually the other branches’ conditionals — compute them as a free self-check

Common Mistakes to Avoid

Key Concept Summary

What You Must Remember

1. Bayes = one branch ÷ branch sum: numerator 9/120, denominator 18/120, answer 1/2
2. Common denominator 120 turns four awkward products into units 9, 8, 1, 0 — count, then divide
3. Partition check first: 3/10 + 1/5 + 1/10 + 2/5 = 1 legitimises total probability
4. The zero late-probability of ‘other’ simply removes its units from the denominator
5. Distractor forensics: 4/9 = P(Bus | Late) and 1/18 = P(Car | Late) — computing them verifies the setup

The Golden Rule for Bayes Reversals

“Causes first, effect second: multiply along each branch, add for the total, then divide your branch by that total — Bayes is bookkeeping, not philosophy.”

Frequently Asked Questions

Q1: What exactly does P(Train | Late) represent?

A: It is the probability that the cause was ‘travelled by train’ after observing the effect ‘reached late’. Formally it is the long-run fraction of late-arriving candidates who had taken the train — a reversal of the supplied information, which was lateness rates for each mode.

Q2: Why do we divide by P(Late) in Bayes’ theorem?

A: Conditioning on ‘late’ restricts attention to only the late-arriving sub-population. P(Late) is the total size of that sub-population (as a probability mass), and the train-late joint probability is the part of it that came by train. The ratio of part to whole is the conditional probability.

Q3: How would the answer change if travelling by other means also had a late probability, say 1/5?

A: The ‘other’ branch would then contribute (2/5)(1/5) = 2/25 = 9.6/120-style mass — exactly 16/120 units — making P(Late) = 34/120 and P(Train | Late) = 9/34. The zero branch is what keeps the denominator at 18 in the original question; any nonzero late rate for other means immediately reshapes the answer.

Q4: Can Bayes’ theorem be applied without the four probabilities summing to 1?

A: The modes must exhaust all possibilities and be mutually exclusive — i.e. form a partition with probabilities summing to 1 — for total probability to be valid. Here 3/10 + 1/5 + 1/10 + 2/5 = 1, so the setup is legitimate; always run this 5-second check before applying Bayes.

Q5: What is the fastest exam-time format for such questions?

A: Draw the four-branch tree, put both rows of given fractions on the branches, multiply along each root-to-late path over a common denominator, add the units, and divide your branch’s units by the total. With denominator 120 bookkeeping the entire question takes about 90 seconds.

Prerequisites to Solve This Question

Before attempting this problem, you should be comfortable with:

1. Conditional probability definition: P(A | B) = P(A ∩ B)/P(B) — Bayes is built from it
2. Law of total probability: partitioning the sample space and summing branch contributions
3. Bayes’ theorem statement: the reversal formula and what each factor means
4. Fraction arithmetic: products like (3/10)(1/4) and LCM-based common denominators (120)
5. Tree diagrams: two-layer trees with branch multiplication as the standard visual

After Solving This, You Can:

• Solve any 3–5 branch Bayes question via common-denominator unit bookkeeping
• Compute the ‘other’ conditionals (P(Bus | Late), P(Car | Late)) as instant distractor checks
• Handle disease-test and machine-defect Bayes problems with the identical tree routine
• Set up reverse-engineering versions: given the posterior, find a missing prior or likelihood
• Distinguish P(A | B) from P(B | A) and P(A ∩ B) on sight — the three-way confusion killer

Study Tips for This Topic

For JEE Main:

1. Tabulate before multiplying: write the two given rows (mode probabilities, late rates) as a table with ‘respectively’ applied — pairing errors then become impossible
2. Pick the LCM early: 120 here unifies everything; unit-counting beats fraction-juggling every time
3. Budget 90 seconds: table (20s) + four products (30s) + sum and divide (30s) + distractor cross-check (10s)

Common JEE Variants:

• Disease testing: rare disease with imperfect test sensitivity/specificity — the classic false-positive shocker
• Machine–factory versions: three machines with output shares and defect rates — find the source of a defective item
• “Given the candidate was on time, find P(train)” — same tree, complement branch (on-time rates 3/4, 2/3, 11/12, 1)
• Two-stage Bayes: posterior from stage 1 becomes the prior for stage 2
• Inverse design: given P(Train | Late) = 1/2, find the missing late-rate — algebra on the unit equation

Difficulty Rating & Exam Frequency

Difficulty Level: ⭐⭐⭐ (3/5) – Medium (pure bookkeeping, but four branches, a zero branch and ‘respectively’ pairing leave room for slips)

JEE Main Frequency: Very High — Bayes reversals with travel/test/machine partitions are a JEE Main staple in almost every session

JEE Advanced Frequency: Medium — Advanced extends to sequential conditioning and posterior-prior chains

Topic Importance: Very High — conditional probability and Bayes form the applied core of the Probability chapter

Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 2 Sep 2026 ·
Question Source: JEE Main 2026 (8 April – Evening Shift) Previous Year Question ·
Topic: Probability – Bayes Theorem with a Four-Mode Partition