ISS 2016 Statistics Paper-1 Solution: Question 25 (Bayes Theorem in QC Testing)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-1 previous year papers, today we pick up right where the last post left off and tackle Question 25 from the 2016 paper. This one sits in the probability section but is really a Bayes' Theorem problem dressed up as a real-world quality control scenario, and it trips up a lot of first-time solvers because of how the conditional probabilities are worded.

Quick Summary

• Topic: Bayes' Theorem / conditional probability in a quality control (QC) testing setup

• Question reference: ISS 2016, Statistics Paper-1, Question 25

• Correct answer: Option (c), 0.28

The Question, As Asked in the Paper

A given lot of items has 2% defectives. A Quality Control (QC) procedure, which is not totally reliable, has the following features:

P[Testing QC procedure indicates an item tested as good | item is indeed good] = 0.95

P[Testing QC procedure indicates an item tested as defective | item is indeed defective] = 0.94

What is the probability that the item tested is defective given that the testing QC procedure indicates it to be defective?

• (a) 0.028

• (b) 0.062

• (c) 0.28

• (d) 0.62

Step 1: Naming the Events Clearly

Before touching any formula, the single most useful habit is to write down, in plain words, exactly what each event stands for. Half the mistakes on this type of question happen because a student confuses "probability the item is defective" with "probability the test says defective." Let's fix that immediately.

Let:

• D = the item is actually defective

• G = the item is actually good (not defective)

• T+ = the QC test indicates "defective"

• T− = the QC test indicates "good"

From the problem, since 2% of the lot is defective:

P(D) = 0.02, and therefore P(G) = 1 − 0.02 = 0.98

Step 2: Translating the Given Conditional Probabilities

The paper gives us two pieces of information, and both need to be read very carefully because of the direction of conditioning:

P(T− | G) = 0.95 — this says: given that an item is genuinely good, the test correctly says "good" 95% of the time.

P(T+ | D) = 0.94 — this says: given that an item is genuinely defective, the test correctly says "defective" 94% of the time. This is the test's sensitivity.

Since "good" and "defective" are the only two outcomes the test can report for a truly good item, P(T+ | G) and P(T− | G) must add up to 1. So:

P(T+ | G) = 1 − P(T− | G) = 1 − 0.95 = 0.05

This 0.05 is the false positive rate — the chance the test wrongly flags a genuinely good item as defective.

Step 3: What the Question Is Actually Asking

We are asked for P(D | T+): given that the test says "defective," what is the actual probability the item is defective? Notice this is the reverse direction from what we were given (we were given P(T+ | D), not P(D | T+)). Whenever a question flips the direction of conditioning like this, that is your signal to reach for Bayes' Theorem.

Bayes' Theorem states:

P(D | T+) = [P(T+ | D) × P(D)] / [P(T+ | D) × P(D) + P(T+ | G) × P(G)]

The denominator here is just the total probability of the test showing "defective," broken down (via the law of total probability) into the two ways that can happen: a truly defective item correctly flagged, or a truly good item wrongly flagged.

Step 4: Plugging in the Numbers

Numerator: P(T+ | D) × P(D) = 0.94 × 0.02 = 0.0188

This is the probability of the joint event "item is defective AND test says defective."

Second term of denominator: P(T+ | G) × P(G) = 0.05 × 0.98 = 0.0490

This is the probability of the joint event "item is good AND test says defective" (a false alarm).

Denominator (total probability of a positive test): 0.0188 + 0.0490 = 0.0678

Now divide:

P(D | T+) = 0.0188 / 0.0678 = 0.2773 (approximately)

Step 5: Matching to the Given Options

0.2773 rounds to 0.28, which matches option (c) exactly. This confirms our answer against the official ISS 2016 Statistics Paper-1 answer key, which also lists option (c) as correct for Question 25.

Pro Tip: The trap in this question is not the arithmetic — it's misreading which conditional probability you were given versus which one you need to find. Students often mistakenly plug P(T+ | D) = 0.94 straight in as the final answer, without realizing the question asks for P(D | T+), the reverse conditioning. The way this kind of question is usually taught is to force yourself to write "P(? | ?)" in words first, for both what's given and what's asked, before writing a single formula. Once you make that a reflex, questions like this stop being tricky and become mechanical.

Final Answer

The correct answer is (c) 0.28, which is the probability that an item is actually defective given that the QC test flagged it as defective. Despite the test being 94% accurate on truly defective items, the low base rate of defectives (only 2%) combined with a 5% false-positive rate means that fewer than 1 in 3 positive test results actually correspond to a genuinely defective item.

Why This Question Matters

This question is a textbook illustration of why raw test accuracy can be misleading when the underlying event (here, a defective item) is rare. It is essentially the same logic behind medical diagnostic testing, spam filters, and fraud detection systems — a highly "accurate" test can still produce a majority of false alarms if the condition being tested for is uncommon. For the ISS exam specifically, Bayes' Theorem questions built around a real-world narrative (QC testing, disease screening, weather forecasting) appear almost every year in Paper-1, so being able to set up the event definitions and the total probability denominator quickly, without hesitation, is a skill worth drilling until it becomes automatic — which is exactly the kind of repeated, structured practice that shows up in a good MCQ-focused revision routine.

Frequently Asked Questions

Why isn't the answer simply 0.94, since that's the test's accuracy on defective items?

0.94 is P(T+ | D), the probability the test says "defective" given the item truly is defective. The question asks for the reverse: P(D | T+), the probability the item is truly defective given the test said so. These are different quantities unless the base rates of "good" and "defective" happen to be equal, which they are not here.

What is the intuitive reason the final probability is so much lower than 0.94?

Because defective items are rare (only 2% of the lot), even a small 5% false-positive rate among the much larger pool of good items (98% of the lot) generates a large number of false alarms. These false alarms end up outnumbering the true positives, pulling the final probability down.

Is this the same as the classic "rare disease" Bayes' Theorem problem?

Yes, structurally it is identical. Replace "defective item" with "has the disease" and "QC test" with "diagnostic test," and the computation is exactly the same. Recognizing this pattern lets you reuse the same four-step setup across many different question wordings.

Do I need to memorize the full Bayes' Theorem formula, or can I derive it each time?

It helps to know the formula, but understanding it as "true positives divided by all positives" (true positives plus false positives) is more reliable under exam pressure than memorizing symbols, since it lets you rebuild the formula from scratch even if you blank on the notation.

What's a quick way to sanity-check this kind of answer during the exam?

Check that your final probability is smaller than the "given" conditional probability whenever the event you're testing for is rarer than its complement — here, defectives (2%) are far rarer than good items (98%), so P(D | T+) should indeed come out noticeably smaller than P(T+ | D) = 0.94, which it does (0.28 versus 0.94).

Where else in the ISS syllabus does this same idea reappear?

The identical Bayes' Theorem structure resurfaces in topics like sensitivity and specificity in biostatistics, screening test evaluation, and even in some estimation and testing of hypotheses questions in Paper-2, so mastering the event-naming habit here pays off well beyond this single question.

If any step above felt unclear, drop a comment below and we'll work through it together — and if you found this helpful, do share it with a fellow ISS aspirant who might be stuck on the same question.