ISS 2017 Statistics Paper-1 Solution: Question 1 (Bayes' Theorem on Exam Guessing)

Welcome back to the ISS Statistics Paper-1 previous-year question series, where we go through one official question at a time, worked out from first principles. Having wrapped up all 80 questions of the 2016 paper, we now move into the 2017 Statistics Paper-1 and start fresh with Question 1 — a classic application of Bayes’ theorem dressed up as an exam-guessing story.

Quick Summary

• Topic:Conditional Probability and Bayes’ Theorem

• Question Reference:ISS 2017, Statistics Paper-1, Question 1

• Correct Answer:Option (d) — 0.94

The Question, As Asked

Leela has appeared in an examination which follows multiple choice questions, each having five possible answers. The probability that she knows an answer is 0.75. If she does not know an answer, she will guess, with the conditional probability 1⁄5 of being correct. The conditional probability that Leela knows the answer, given that she gives the correct answer, is:

• (a) 0.25

• (b) 0.80

• (c) 0.90

• (d) 0.94

Step 1: Identify What Is Given and What Is Asked

Before touching any formula, it helps to write down, in plain words, exactly what information the question hands you and exactly what it wants back. That one habit prevents half the silly mistakes people make under exam pressure.

Let us define two events:

• K: the event that Leela knows the answer.

• K′(read as “not K”): the event that Leela does not know the answer.

From the question:

• P(K) = 0.75 — the probability she knows the answer.

• P(K′) = 1 − 0.75 = 0.25 — the probability she does not know the answer (since K and K′ are complementary, their probabilities must add up to 1).

Next, define a third event:

• C: the event that Leela’s given answer turns out to be correct.

The question tells us two conditional probabilities involving C:

• If Leela knows the answer, she is certainly correct, so P(C | K) = 1.

• If Leela does not know the answer, she guesses, and is told to be correct with conditional probability 1⁄5, so P(C | K′) = 1⁄5 = 0.2.

And finally, the question asks for P(K | C) — the probability that Leela knew the answer,given that she got it right. Notice this is the reverse of the conditional probabilities we were handed: we were given P(C | K) and P(C | K′), but we are asked to find P(K | C). Whenever a question gives you probabilities in one direction (effect given cause) and asks for the reverse direction (cause given effect), that is your signal to reach for Bayes’ theorem.

Step 2: Write Down Bayes’ Theorem

Bayes’ theorem, in the two-event form we need here, states:

P(K | C) = [ P(C | K) × P(K) ] ÷ P(C)

Here, P(C) is the overall (unconditional) probability that Leela answers correctly, regardless of whether she knew the answer or guessed it. We do not know this number directly, but we can build it from the two cases we already have, using the law of total probability.

Step 3: Compute P(C) Using the Law of Total Probability

Since K and K′ are complementary events (together they cover every possibility — either she knows it or she doesn’t, there is no third option), the event C can be split into two mutually exclusive pieces:

P(C) = P(C | K) × P(K) + P(C | K′) × P(K′)

This formula simply says: “the chance of being correct overall equals (the chance of knowing it times the chance of being correct when you know it) plus (the chance of not knowing it times the chance of being correct when you guess).” Substituting the numbers:

P(C) = (1 × 0.75) + (0.2 × 0.25)

P(C) = 0.75 + 0.05

P(C) = 0.80

So, out of all the questions Leela attempts, she ends up correct on 80% of them on average — some because she genuinely knew the answer, and a smaller slice because she got lucky guessing.

Pro Tip: The step students most often fumble here is forgetting that P(C) must account for

Step 4: Apply Bayes’ Theorem to Find P(K | C)

Now that we have every piece, we substitute back into the Bayes’ theorem formula from Step 2:

P(K | C) = [ P(C | K) × P(K) ] ÷ P(C)

P(K | C) = (1 × 0.75) ÷ 0.80

P(K | C) = 0.75 ÷ 0.80

To simplify this division cleanly, multiply numerator and denominator by 100 to clear the decimals:

P(K | C) = 75 ÷ 80

Both 75 and 80 are divisible by 5:

P(K | C) = 15 ÷ 16

Carrying out this division: 15 ÷ 16 = 0.9375.

Rounded to two decimal places, as the answer options are given, this is 0.94.

Final Answer and Verification Against the Official Key

Our independently derived answer is 0.94, which corresponds to option (d). This matches the official ISS 2017 Statistics Paper-1 answer key exactly, which also lists option (d) as correct for Question 1. No discrepancy here — our step-by-step derivation and the official key are in full agreement.

Why This Question Matters

This question is a textbook example of why Bayes’ theorem exists at all: our natural intuition handles “if she knows it, what’s the chance she’s right” easily, but “if she’s right, what’s the chance she actually knew it” is a genuinely different, harder question that needs a formal tool to answer correctly. This exact knows-vs-guesses setup reappears constantly in the ISS syllabus, in disease-testing problems, quality-control problems, and multiple-source classification problems — anywhere you need to reason backward from an observed outcome to its probable cause. Mastering the tree-diagram approach here pays off across dozens of future questions, not just this one.

Frequently Asked Questions

Why is P(C | K) equal to 1 and not something less than 1?

The question states that if Leela knows the answer, she simply writes it down correctly. There is no uncertainty in that branch at all — knowing the answer and giving the correct answer are treated as the same event here, so the conditional probability is exactly 1, the maximum possible value.

Where does the number 1/5 actually come from, given that each question has five options?

The question explicitly states the guessing probability is 1/5, and this happens to align neatly with there being five answer choices per question (a pure random guess among five equally likely options would succeed one time in five). Always use the number given in the question text itself rather than assuming it from the option count, since some questions deliberately set a different guessing probability.

Is Bayes’ theorem the only way to solve this, or could a frequency/table method work too?

Yes, a frequency-table (natural frequency) approach works just as well and can be more intuitive for some students. Imagine 10,000 students: 7,500 know the answer and all 7,500 are correct; of the remaining 2,500 who guess, 20% (500) are correct. Out of the 8,000 total correct answers, 7,500 actually knew it, giving 7,500/8,000 = 0.9375, the same result.

What is the most common calculation mistake on this type of question?

The most frequent error is computing only the numerator, P(C | K) × P(K) = 0.75, and mistaking that for the final answer, forgetting to divide by the total probability of being correct, P(C). That gives 0.75 instead of 0.9375, which does not even appear among the answer choices — always a useful self-check that something was skipped.

Does the order in which K and K′ are considered matter for the final answer?

No. Whether you first compute the “knows it” branch or the “guesses it” branch, the law of total probability simply adds the two contributions, and addition is commutative. What matters is that both branches are included and none are double-counted.

How should this type of question be practiced for the actual exam?

Drill several Bayes’ theorem variants (medical testing, quality control, multiple-machine production) until the tree-diagram structure becomes automatic rather than something you have to re-derive each time; under a strict two-hour, 80-question time limit, recognizing the pattern instantly is what actually saves time, far more than knowing the formula itself.

That wraps up Question 1 of the 2017 Statistics Paper-1. If anything in the derivation felt unclear, or you solved it a different way and want a second opinion, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.