ISS 2016 Statistics Paper-1 Solution: Question 3 (Binomial Probability of Success)

We're carrying forward our question-by-question walk through the ISS Statistics Paper-1 previous year papers, and today's stop is Question 3 of the 2016 paper. After two questions built around jointly distributed and Bernoulli random variables, this one shifts to the binomial distribution — and it tests something every aspirant needs to have completely automatic: reading a relation between two binomial probabilities and turning it into a clean equation for the probability of success.

Quick Summary

• Topic:Binomial distribution — finding the probability of success from a ratio condition on two probabilities

• Question Reference:ISS 2016, Statistics Paper-1, Question 3

• Correct Answer:(b) 1/4

The Question

If in 6 trials, X is a binomial variate which follows the relation 9P(X = 4) = P(X = 2), then what is the probability of success?

• (a) 1/8

• (b) 1/4

• (c) 3/8

• (d) 3/4

Step 1: Write Down the Binomial Probability Formula

X is given to be a binomial variate based on n = 6 trials, with probability of success p (and probability of failure q = 1 − p). For a binomial distribution, the probability of getting exactly k successes in n trials is:

P(X = k) = nCk pk (1 − p)n − k

Here n = 6 is fixed by the problem. We need two specific values of this formula: P(X = 4) and P(X = 2).

Step 2: Write Out P(X = 4) and P(X = 2) Explicitly

Substituting k = 4 and n = 6:

P(X = 4) = 6C4 p4 (1 − p)6 − 4 = 6C4 p4 (1 − p)2

Substituting k = 2 and n = 6:

P(X = 2) = 6C2 p2 (1 − p)6 − 2 = 6C2 p2 (1 − p)4

Now we need the actual numeric values of the two combination terms. Recall that nCk = n! / (k!(n − k)!).

6C4 = 6! / (4! × 2!) = (6 × 5 × 4 × 3 × 2 × 1) / ((4 × 3 × 2 × 1) × (2 × 1)) = 720 / (24 × 2) = 720 / 48 = 15

6C2 = 6! / (2! × 4!) = 720 / (2 × 24) = 720 / 48 = 15

So it turns out 6C4 and 6C2 are both equal to 15. This is not a coincidence — it's a direct consequence of the symmetry property nCk = nCn−k, and here 6 − 4 = 2, so 6C4 = 6C2 automatically. Keeping this symmetry in mind saves you a full factorial calculation.

So we now have:

P(X = 4) = 15 p4 (1 − p)2

P(X = 2) = 15 p2 (1 − p)4

Step 3: Substitute Into the Given Relation

The question tells us that 9P(X = 4) = P(X = 2). Substituting the expressions we just found:

9 × 15 p4 (1 − p)2 = 15 p2 (1 − p)4

Since 15 appears on both sides, it cancels immediately, leaving:

9 p4 (1 − p)2 = p2 (1 − p)4

Step 4: Simplify by Cancelling Common Factors

Both sides contain p2 and (1 − p)2 as common factors (since 0 < p < 1, neither p nor (1 − p) is zero, so it is completely safe to divide both sides by p2(1 − p)2). Dividing:

9 p4 (1 − p)2 / [p2 (1 − p)2] = p2 (1 − p)4 / [p2 (1 − p)2]

9 p2 = (1 − p)2

Pro Tip: This is the exact step where most students lose marks under exam pressure — they take the square root of both sides and forget that √(x²) = |x|, not just x. Since 9p² = (1 − p)² can be rewritten as (3p)² = (1 − p)², taking square roots gives 3p = ±(1 − p). It's only because we know 0 < p < 1 (so both 3p and 1 − p are positive here) that we can safely discard the negative root. This habit — pausing to check the valid range before picking a root — is exactly the kind of thing that gets drilled repeatedly in a good coaching classroom until it becomes second nature, because in later chapters (like MLE derivations or variance-stabilizing transformations) picking the wrong root silently gives you a completely wrong final answer with no error message to warn you.

Step 5: Solve the Resulting Linear Equation

Taking the square root of both sides of 9p² = (1 − p)²:

√(9p²) = √((1 − p)²)

3p = |1 − p|

Since 0 ≤ p ≤ 1 (p is a probability), we have 1 − p ≥ 0, so |1 − p| = 1 − p. This gives the valid equation:

3p = 1 − p

Now solve for p by collecting like terms. Add p to both sides:

3p + p = 1

4p = 1

Divide both sides by 4:

p = 1/4

Step 6: Sanity-Check the Answer

It's always worth plugging the answer back into the original relation to make sure nothing was mishandled. With p = 1/4, we have 1 − p = 3/4.

P(X = 4) = 15 × (1/4)4 × (3/4)2 = 15 × (1/256) × (9/16) = 135/4096

P(X = 2) = 15 × (1/4)2 × (3/4)4 = 15 × (1/16) × (81/256) = 1215/4096

Check: 9 × P(X = 4) = 9 × 135/4096 = 1215/4096, which is exactly equal to P(X = 2) = 1215/4096. The relation holds exactly, confirming p = 1/4 is the correct probability of success.

Final Answer

The probability of success is p = 1/4, which is option (b). This matches the official ISS 2016 Statistics Paper-1 answer key, which also lists the answer to Question 3 as (b).

Why This Question Matters

On the surface this looks like a routine binomial probability question, but it is really testing three separate skills at once: knowing the binomial PMF formula cold, recognizing and exploiting the symmetry of combinations (nCk = nCn−k) to save calculation time, and — most importantly — handling square roots of squared expressions correctly by respecting the valid domain of the variable. Questions structured exactly like this one show up repeatedly across ISS and similar statistics services exams, just with different numbers, so mastering the method here (rather than memorizing this specific answer) pays off directly on test day.

Frequently Asked Questions

Why are C(6,4) and C(6,2) equal in this problem?

This follows from the general combinatorial identity nCk = nCn−k, which holds because choosing k items to include from a set of n is equivalent to choosing n − k items to exclude. Here n = 6 and 6 − 4 = 2, so 6C4 automatically equals 6C2, both being 15.

Why can we ignore the negative root when solving 9p² = (1 − p)²?

Because p represents a probability, it must satisfy 0 ≤ p ≤ 1, which forces 1 − p to be non-negative. When you take a square root, √(x²) = |x|, and here the negative root (3p = p − 1) would force p to be negative, which is impossible for a probability. So only the positive branch, 3p = 1 − p, is valid.

Is there a faster way to solve this without expanding the full binomial formula?

Yes — since P(X = k) = nCk pk(1 − p)n−k, the ratio P(X = 4)/P(X = 2) simplifies directly to (6C4/6C2) × (p/(1−p))², and because the combination terms are equal here, the ratio reduces immediately to (p/(1−p))². Setting this ratio equal to 1/9 (from 9P(X=4) = P(X=2)) gets you to the same equation with less writing.

What if the combination terms in the numerator and denominator hadn't been equal?

You would simply carry the ratio of the two combination values as a constant multiplier alongside the p and (1 − p) powers, rather than having it cancel away. The rest of the method — isolating the ratio p/(1−p), taking a square root if needed, and solving the resulting linear or quadratic equation — stays exactly the same.

How is this different from a Poisson or normal approximation type question?

This question uses the exact binomial PMF with a fixed, small number of trials (n = 6), so there's no need for any approximation — you work with the exact formula throughout. Poisson or normal approximations only become relevant when n is large and exact binomial computation becomes impractical, which is a separate topic tested elsewhere in the syllabus.

What's the most common calculation mistake in this type of question?

Beyond the root-sign issue discussed above, students frequently make an arithmetic slip while expanding the factorials for nCk, or forget to write (1 − p) correctly raised to the (n − k) power. Writing out each step of the combination and power calculation separately, as done above, is the most reliable way to avoid both errors.

If any step above wasn't clear, or you'd like to see this solved a different way, drop a comment below — and if you know a fellow ISS aspirant working through the same papers, feel free to share this post with them.