ISS 2016 Statistics Paper-1 Solution: Question 10 (Expected Unique Elements in Sampling With Replacement)
Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 10 of the 2016 paper. If you've been following this series from Question 1 onward, this one is a nice change of pace — it moves us from pure probability distributions into a classic sampling theory result that shows up again and again in later papers.
Quick Summary
Topic:Simple random sampling with replacement — expected number of distinct (unique) population elements captured in the sample
Question Reference:ISS 2016, Statistics Paper-1, Question 10
Correct Answer:Option (b) — 100 × [1 − (99/100)10]
The Question, As Asked
A simple random sample of size 10 is selected with replacement from a population of size 100. What is the expected number of unique elements in the sample?
100 × (99/100)10
100 × [1 − (99/100)10]
100 × (9/10)10
100 × (1 − 99/100)10
Step 1: Understand What "With Replacement" Really Means
The population has N = 100 distinct units — think of them as 100 numbered chits in a box. We draw n = 10 chits one at a time, but after every draw we put the chit back before drawing again. Because the chit goes back in, the exact same unit can be picked more than once, and some units may never get picked at all in 10 draws.
The question asks for the expected number of distinct units that show up at least once in our sample of 10 draws. This is different from asking "how many draws were there" (that's just 10, always) — it's asking how many of the 100 original units got "touched" by at least one of those 10 draws.
Step 2: Set Up an Indicator Variable for Each Population Unit
This is the single most useful trick in this whole problem, and it's worth learning properly because it reappears in many sampling and occupancy-type questions. Instead of trying to directly count how many unique units appear, we build one indicator variable per unit and add them up.
For each unit i (i = 1, 2, ..., 100) in the population, define:
Ii = 1, if unit i is selected at least once in the 10 draws Ii = 0, if unit i is never selected in the 10 draws
Now here is the key identity: the total number of unique elements in the sample, call it U, is simply
U = I1 + I2 + I3 + ... + I100
This is true because each Ii contributes exactly 1 to the sum if and only if unit i was picked at least once — so the sum literally counts how many distinct units appear. This is called the "indicator variable" or "linearity of expectation" method.
Pro Tip: The trickiest part of this question is trusting that you can take the expectation of a sum of 100 dependent-looking indicator variables just by adding up 100 individual expectations, without worrying about dependence between them at all. Linearity of expectation, E(I1 + I2 + ... + I100) = E(I1) + E(I2) + ... + E(I100), holds no matter whether the individual indicators are independent of each other or not. The way this identity is usually drilled into students is by first solving small toy versions by brute force (say N = 3, n = 2) and checking the shortcut against direct enumeration — once you've verified it that way a couple of times, you stop needing to re-derive it and can just apply it directly on exam day.
Step 3: Find the Expectation of a Single Indicator Variable
Since Ii only takes the values 0 or 1, it is a Bernoulli-type variable, and for any such variable:
E(Ii) = 1 × P(Ii = 1) + 0 × P(Ii = 0) = P(Ii = 1) = P(unit i is selected at least once in 10 draws)
So we need P(unit i is picked at least once). It is much easier to first find the complementary probability — the probability that unit i is never picked in all 10 draws — and then subtract from 1.
On any single draw (with replacement, so every draw is independent and each of the 100 units is equally likely):
P(unit i is NOT picked on one particular draw) = 99/100
because there are 99 other units out of 100 that could be picked instead of unit i. Since the 10 draws are independent of each other (that is exactly what "with replacement" buys us — the population composition never changes between draws), the probability that unit i is missed on all 10 draws is the product of 10 independent "miss" probabilities:
P(unit i never selected in 10 draws) = (99/100)10
Therefore:
P(unit i selected at least once) = 1 − (99/100)10
So E(Ii) = 1 − (99/100)10 for every single i, and this value is exactly the same for all 100 units, since every unit is symmetric — none of them is treated specially by the sampling scheme.
Step 4: Add Up All 100 Expectations
Now we bring Step 2 and Step 3 together. Using linearity of expectation:
E(U) = E(I1) + E(I2) + ... + E(I100)
Since every one of the 100 terms equals the same value 1 − (99/100)10, this sum is just 100 copies of that value added together, which is the same as multiplying:
E(U) = 100 × [1 − (99/100)10]
Step 5: Match With the Given Options and Sanity-Check Numerically
Comparing with the four choices, this is exactly option (b): 100 × [1 − (99/100)10].
It's always worth a quick numerical sanity check before finalizing an answer. Computing 0.99 raised to the power 10:
0.9910 ≈ 0.9044
So 1 − 0.9044 = 0.0956, and 100 × 0.0956 ≈ 9.56.
Does this number make intuitive sense? We drew 10 chits (with repetition allowed) out of 100. Since 100 is much larger than 10, repeats should be fairly rare, so the number of unique units should be a little less than 10 — and 9.56 fits that intuition perfectly. If your formula had instead given you something like 90 or 0.1, that would have been an immediate red flag to go back and recheck the setup.
Let's also rule out the other options quickly, since a good MCQ strategy is understanding why the wrong options are wrong, not just why the right one is right:
Option (a), 100 × (99/100)10 ≈ 90.44, is actually the expected number of units never selected — the complement of what was asked.
Option (c), 100 × (9/10)10, incorrectly uses 9/10 instead of 99/100 — this would be the answer if the population size were 10, not 100.
Option (d), 100 × (1 − 99/100)10, incorrectly raises the small quantity (1 − 99/100) = 1/100 to the 10th power instead of raising 99/100 to the 10th power and then subtracting from 1 — a subtle but very different expression algebraically.
Final Answer: Option (b) — 100 × [1 − (99/100)10], which agrees with the official answer key (B) for Question 10.
Why This Question Matters
This question is a gateway into a whole family of "occupancy" and "coverage" problems in sampling theory — the same indicator-variable technique is what's used to derive results like the expected number of empty boxes in a balls-and-boxes problem, the expected number of distinct coupons collected in a coupon-collector-style setup, or the expected overlap between two random samples. Once this method is internalized, several otherwise-intimidating combinatorial expectation questions in later ISS papers become routine applications of the same three steps: define an indicator, find its expectation using a complement, and sum by linearity.
Frequently Asked Questions
Why does using indicator variables make this problem so much easier than trying to count unique elements directly?
Counting unique elements directly would require working out the full probability distribution of "how many distinct units appear in 10 draws," which is a genuinely hard combinatorial calculation involving inclusion-exclusion over 100 units. By instead expressing the count as a sum of 0/1 indicators and using linearity of expectation, we completely bypass needing that distribution — we only ever need the expectation of one simple 0/1 variable, repeated 100 times.
Does linearity of expectation require the indicator variables to be independent?
No, and this is the most important thing to remember from this question. E(X + Y) = E(X) + E(Y) is always true, for any random variables X and Y, whether or not they are independent or even correlated. Independence would matter if we were trying to find the variance of the sum, but for the expectation of a sum, it is never required.
What changes in this formula if the sampling is done without replacement instead?
Without replacement, every draw would automatically be a distinct unit (since a unit can't be picked twice), so the expected number of unique elements would simply equal the sample size, n = 10, with probability 1. The interesting probability calculation in this question exists only because replacement is allowed.
How would the answer change if the sample size were different, say n = 20 instead of 10?
The same formula would apply with the exponent changed: E(U) = 100 × [1 − (99/100)20]. As n grows larger, (99/100)n shrinks toward zero, so E(U) approaches 100 — meaning with enough draws, you'd expect to eventually touch nearly every unit in the population at least once.
Is this the same idea as the classic "birthday problem"?
It's a close cousin. The birthday problem usually asks for the probability that at least one collision (repeat) occurs among a group of people, whereas this question asks for the expected count of units that are hit at least once. Both problems rely on the same complement trick — computing the probability of "no hit" or "no collision" first, then working from there — applied to slightly different quantities.
What's a good way to double check this kind of answer under exam time pressure?
Plug in small, easy numbers mentally as a sanity check, exactly as done in Step 5 above: with a large population and a small sample, the expected number of unique elements should be just slightly less than the sample size, never more than it and never dramatically smaller. If your final expression doesn't roughly satisfy that check, it's worth re-reading the question to see if you've mixed up N and n somewhere.
If any step above still feels unclear, or if you spot a different way to approach this question, drop a comment below — working through the disagreement is often where the real learning happens. And if you found this useful, do share it with a fellow ISS aspirant who might be stuck on the same question.
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