ISS 2016 Statistics Paper-1 Solution: Question 19 (Sum of Independent Binomials)
Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we land on Question 19 from the 2016 paper. It looks like a short "which statements are true" question, but it hides one of the most useful facts in the whole binomial-distribution toolkit, so it is worth slowing down and working through every line by hand.
Quick Summary
Topic: Independent Binomial random variables, their sum, and conditioning on the sum
Question Reference: ISS 2016, Statistics Paper-1, Question 19
Correct Answer: Option (c) — Statements 1 and 2 only
The Question, Exactly As Asked
If X and Y are two independent random variables with Binomial distributions having integer parameters m and n and the same probability parameter p, then consider the following statements:
m + n − X − Y has a Binomial distribution.
The conditional distribution of X given the sum X + Y is a hyper-geometric distribution.
The conditional distribution of X given the sum X + Y is again a Binomial distribution.
Which of the above statements is/are correct?
(a) 1 only
(b) 2 only
(c) 1 and 2 only
(d) 1 and 3 only
Step 1: Set Up What We Actually Know About X and Y
We are told X ~ Binomial(m, p) and Y ~ Binomial(n, p), and X, Y are independent. The key detail — easy to skim past — is that both variables share the exact same success probability p. Everything in this question hinges on that shared p.
Recall the probability mass function of a Binomial(n, p) variable:
P(X = x) = C(m, x) · px · qm−x, for x = 0, 1, 2, …, m, where q = 1 − p.
Step 2: Prove X + Y Is Also Binomial (This Is the Building Block)
Before touching statement 1, we need a fact that the paper doesn't spell out but expects you to know: if X ~ Binomial(m, p) and Y ~ Binomial(n, p) are independent with the same p, then X + Y ~ Binomial(m + n, p).
Here is the plain-language reason, no moment generating functions required. Binomial(m, p) is just "count the successes in m independent coin tosses, each with success probability p." Binomial(n, p) is "count the successes in n more independent tosses of the same coin." If you glue these two batches of tosses together, you now have m + n independent tosses of the same coin, and X + Y simply counts the total successes across all of them. Counting successes in m + n independent p-coin tosses is, by definition, Binomial(m + n, p). So:
X + Y ~ Binomial(m + n, p)
Step 3: Check Statement 1 — Is m + n − X − Y Binomial?
Let Z = X + Y, so Z ~ Binomial(m + n, p) from Step 2. Statement 1 asks about m + n − Z.
Think about what Z counts: successes in m + n tosses. Then (m + n) − Z counts everything that was not a success in those same m + n tosses — in other words, the number of failures. If each toss succeeds with probability p, it fails with probability q = 1 − p, and the failures are still independent across the m + n tosses. Counting failures in m + n independent trials is exactly Binomial(m + n, q), i.e. Binomial(m + n, 1 − p).
So m + n − X − Y ~ Binomial(m + n, 1 − p), which is indeed a Binomial distribution (just with success probability flipped to q instead of p).Statement 1 is TRUE.
Step 4: Find the Conditional Distribution of X Given X + Y = k
This is the heart of the question, and where statements 2 and 3 are decided. We want P(X = x | X + Y = k) for a fixed total k.
By the definition of conditional probability:
P(X = x | X + Y = k) = P(X = x and X + Y = k) / P(X + Y = k)
Because X and Y are independent, the event "X = x and X + Y = k" is the same as "X = x and Y = k − x", and its probability is just the product P(X = x) · P(Y = k − x). So:
P(X = x | X + Y = k) = [P(X = x) · P(Y = k − x)] / P(X + Y = k)
Now substitute the Binomial formulas for each piece. Using X ~ Binomial(m, p), Y ~ Binomial(n, p), and Z = X + Y ~ Binomial(m + n, p) from Step 2:
Numerator = C(m, x) px qm−x × C(n, k−x) pk−x qn−(k−x)
Denominator = C(m+n, k) pk q(m+n)−k
Let's simplify the numerator's powers of p and q first, since that is the part students most often fumble. Adding the exponents of p: x + (k − x) = k. Adding the exponents of q: (m − x) + (n − k + x) = m + n − k. So the numerator becomes:
C(m, x) · C(n, k−x) · pk · q(m+n)−k
Now divide by the denominator. Notice pk · q(m+n)−k appears in both the numerator and the denominator — they cancel completely, leaving:
P(X = x | X + Y = k) = C(m, x) · C(n, k−x) / C(m+n, k)
Pro Tip: That cancellation of pkq(m+n)−k is the entire trick of the question. It only happens because X and Y share the same p — if they had different success probabilities, the p and q powers would not match up and would not cancel, and the conditional distribution would not simplify into a clean closed form at all. Whenever an exam question insists on "the same probability parameter p" for two independent Binomials, that phrase is a signal: it is telling you, in advance, that this cancellation is coming. Students who train themselves to spot that phrase and immediately expect a p-free ratio of combinations tend to solve this entire family of questions — Binomial-given-sum, Poisson-given-sum, and their relatives — almost on sight, which is exactly the kind of pattern recognition this topic is usually drilled around.
Step 5: Recognize the Result As a Hypergeometric Distribution
The expression C(m, x) · C(n, k−x) / C(m+n, k) is not just "a ratio of combinations" — it is precisely the probability mass function of a Hypergeometric distribution. To see why, picture an urn with m red balls and n black balls (total m + n balls). If you draw k balls without replacement, the probability that exactly x of them are red is:
P(x red in the sample) = C(m, x) · C(n, k−x) / C(m+n, k)
That is exactly the formula we derived. So the conditional distribution of X, given X + Y = k, is Hypergeometric with population size m + n, m "successes" in the population, and sample size k.
Statement 2 says this conditional distribution is hypergeometric — that matches exactly what we derived, so Statement 2 is TRUE.
Statement 3 says this conditional distribution is again Binomial — but we just showed it is Hypergeometric, not Binomial (there is no p or q left anywhere in the final formula, which is the clearest sign it cannot be Binomial). So Statement 3 is FALSE.
Step 6: Combine the Results and Match to an Option
Putting Steps 3 and 5 together:
Statement 1 — TRUE
Statement 2 — TRUE
Statement 3 — FALSE
So the correct statements are 1 and 2 only, which is option (c). Checking against the official ISS 2016 Statistics Paper-1 answer key, the recorded answer for Question 19 is also C. Our independently worked derivation matches the official key exactly.
Why This Question Matters
This question is a compact test of three separate ideas at once: the additive property of Binomial distributions with a common p, the "reflection" trick that turns successes into failures, and the classic result that conditioning a sum of independent Binomials on its own total produces a Hypergeometric distribution. That last result is a genuine workhorse in statistical theory — it is the exact same mechanism used to justify Fisher's exact test for 2×2 contingency tables, where you condition on both margins being fixed. Seeing it appear here, in a two-line MCQ, is a preview of a much bigger idea you will meet again in inference courses.
Frequently Asked Questions
Why do X and Y need the same probability parameter p for X + Y to be Binomial?
The additive property relies on being able to describe X and Y as counts of successes from a single shared sequence of independent coin tosses, each with success probability p. If p differed between X and Y, the combined variable would still exist, but it would not reduce to a single clean Binomial(m+n, p) form — the pmf of the sum would instead be a more complicated convolution with no simple closed expression.
What exactly is a Hypergeometric distribution, in plain terms?
It describes the number of "special" items you get when you draw a fixed-size sample without replacement from a finite population that contains a known number of special and non-special items. In this question, once you know the total X + Y = k, you are effectively "sampling" k items without replacement from a population of m + n underlying trials, of which m were the "X-type" trials.
Why does conditioning change the distribution from Binomial to Hypergeometric at all?
Before conditioning, X can independently take any value from 0 to m, with randomness coming from p. Once you fix X + Y = k, the randomness in p disappears entirely — you are left only with the combinatorial question of how the k total successes could have been split between the m "slots" belonging to X and the n slots belonging to Y, and that splitting question is answered by hypergeometric counting.
Is there a quick way to remember which statement combination is correct without doing the full algebra in an exam?
Not a shortcut you should trust blindly, but a sanity check: since the conditional pmf you get after conditioning on the sum has no p or q terms left in it, it cannot be Binomial (which always depends on p). That single observation is often enough to eliminate any option claiming the conditional distribution stays Binomial, which is exactly what kills statement 3 here.
Does this same result work for Poisson random variables too?
Yes, and it is one of the most commonly tested parallels in these exams. If X and Y are independent Poisson variables, X + Y is also Poisson, and the conditional distribution of X given X + Y = k turns out to be Binomial — not Hypergeometric. It is worth carefully comparing the two derivations side by side so you don't mix up which family (Binomial-conditioning or Poisson-conditioning) produces which resulting distribution.
Where do statement-based MCQs like this typically trip students up?
Most errors come from assuming a statement is true just because it "sounds plausible" — for instance, assuming the conditional distribution must stay in the same family as the original variables. The safer approach, as this solution shows, is to always write out the actual pmf ratio and simplify it algebraically rather than guessing from intuition alone.
If any step above felt rushed or you would solve it differently, drop a comment below — working through the disagreement is often where the real learning happens. And if you know a fellow ISS aspirant who is grinding through the same PYQ set, feel free to pass this one along to them too.
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