ISS 2016 Statistics Paper-1 Solution: Question 11 (Expected Empty Boxes in Random Allocation)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 11 of the 2016 paper. This one looks like a simple counting puzzle at first glance, but it is really testing whether you can spot when linearity of expectation lets you skip a much harder direct calculation entirely.

Quick Summary

• Topic: Probability – expectation via linearity, occupancy/balls-into-boxes problem

• Question Reference: ISS 2016, Statistics Paper-1, Question 11

• Correct Answer: Option (c), 910/109

The Question

10 balls are placed in 10 boxes independently at random. Assuming that all 10 boxes were initially empty, what is the expected number of boxes that remain empty?

• (a) (9/10)9

• (b) 99/1010

• (c) 910/109

• (d) (9/10)10

Step 1: Understand What "Placed Independently at Random" Means

We have 10 boxes, numbered 1 through 10. We throw 10 balls, one at a time, and each ball lands in one of the 10 boxes completely at random, with each box equally likely, and every ball's landing spot is independent of every other ball's landing spot. This is exactly the same setup as rolling a fair 10-sided die 10 times and recording which face comes up each time — a box is like a face of the die, and a ball landing in it is like the die showing that face.

So for any single ball, the probability that it lands in a specific box, say Box 1, is 1/10. The probability that it lands in any of the other 9 boxes (that is, it misses Box 1) is 9/10.

Step 2: Find the Probability That One Particular Box Stays Empty

Let us focus on just one box, Box 1, and ask: what is the probability that Box 1 ends up empty after all 10 balls have been thrown?

Box 1 stays empty only if every single one of the 10 balls misses it. Since the balls are placed independently, we can multiply the probability of "this ball misses Box 1" across all 10 balls:

P(Box 1 is empty) = P(ball 1 misses Box 1) × P(ball 2 misses Box 1) × ... × P(ball 10 misses Box 1)

Each of these individual probabilities is 9/10, and there are 10 balls, so:

P(Box 1 is empty) = (9/10) × (9/10) × ... × (9/10) [10 times] = (9/10)10

By exactly the same reasoning, this is true for every box, not just Box 1. Box 2, Box 3, all the way up to Box 10 — each one individually has probability (9/10)10 of remaining empty. The boxes are symmetric; nothing distinguishes one from another before the balls are thrown.

Step 3: Set Up an Indicator Variable for Each Box

Now we want the expected number of boxes that remain empty, not just the probability that one particular box is empty. This is where a very standard and powerful trick comes in: indicator variables.

For each box i (where i runs from 1 to 10), define a random variable:

Ii = 1, if Box i remains empty Ii = 0, if Box i receives at least one ball

The total number of boxes that remain empty, call it N, is simply the sum of all these indicator variables:

N = I1 + I2 + I3 + ... + I10

This is just a formal way of saying "count 1 for every box that turns out empty, and add them all up." That sum is, by definition, the number of empty boxes.

Step 4: Apply Linearity of Expectation

Here is the key idea that makes this problem easy instead of hard. The expectation of a sum of random variables always equals the sum of their individual expectations, and this is true regardless of whether the variables are independent of each other or not. This property is called linearity of expectation, and it is one of the most useful tools in all of probability theory.

So:

E(N) = E(I1) + E(I2) + E(I3) + ... + E(I10)

Now, for any indicator variable (a variable that only takes the values 0 or 1), the expectation is simply the probability that it equals 1. This is a basic fact worth memorizing: if I is an indicator variable, E(I) = 1 × P(I = 1) + 0 × P(I = 0) = P(I = 1).

We already worked out in Step 2 that P(Ii = 1) = P(Box i is empty) = (9/10)10, for every single box i. So each term in the sum equals (9/10)10, and there are 10 terms:

E(N) = 10 × (9/10)10

Pro Tip: The trap in this question is trying to reason about all 10 boxes jointly — asking things like "given that Box 1 is empty, how does that affect Box 2's chance of being empty?" That path leads into messy dependent-events territory, because the boxes are

Step 5: Simplify to Match the Answer Choices

We have E(N) = 10 × (9/10)10. Let's rewrite this so it matches the form of the given options. Recall that (9/10)10 means 910 divided by 1010:

E(N) = 10 × (910 / 1010)

Now, 10 in the numerator can cancel with one factor of 10 in the denominator's 1010, since 1010 = 10 × 109:

E(N) = (10 × 910) / (10 × 109) = 910 / 109

This matches option (c) exactly.

Step 6: Sanity-Check with a Numeric Value

It always helps to plug in numbers and see if the answer feels reasonable. Computing 910 = 3,486,784,401 and 109 = 1,000,000,000, we get:

E(N) = 3,486,784,401 / 1,000,000,000 ≈ 3.487

Does this make sense? With 10 balls thrown into 10 boxes at random, we would intuitively expect a fair number of boxes to end up empty just from randomness — balls will clump into some boxes rather than spreading out perfectly one-per-box (which would happen only along a very small fraction of all possible outcomes). An expected value of roughly 3.5 out of 10 boxes staying empty lines up well with that intuition — not too high, not too low.

Final Answer and Confirmation

The expected number of empty boxes is 910/109, which corresponds to option (c). This matches the official ISS 2016 Statistics Paper-1 answer key, which also lists (c) as the correct choice for Question 11. Our independent derivation via indicator variables and linearity of expectation agrees fully with the key.

Why This Question Matters

This question is a textbook example of the "balls into boxes" or occupancy problem, one of the most recurring themes in probability at the ISS, IIT JAM, and similar exams. The identical technique — define an indicator variable for each unit of interest, compute its expectation via a simple probability, then sum using linearity — solves a huge family of questions: expected number of matches in a shuffled deck, expected number of fixed points in a random permutation, expected number of empty urns, expected number of coupons needed to complete a collection, and more. Recognizing this pattern the moment you see "expected number of X that satisfy some condition" can turn a seemingly complicated combinatorics problem into two or three lines of clean algebra.

Frequently Asked Questions

Why can we use linearity of expectation even though the events "Box 1 is empty" and "Box 2 is empty" are not independent?

Linearity of expectation holds unconditionally for any collection of random variables, whether or not they are independent. It only requires that each individual variable has a well-defined expectation; it says nothing about how the variables relate to each other. This is precisely what makes it such a powerful shortcut in problems like this one, where the events are clearly dependent but we still want their combined expected count.

What is an indicator variable, and why is its expectation just a probability?

An indicator variable is a random variable that takes only the value 1 (if some event happens) or 0 (if it doesn't). Its expectation is calculated as 1 times the probability of the event plus 0 times the probability of the complementary event, which simplifies to just the probability of the event itself. This makes indicator variables extremely convenient for counting problems, since summing them counts how many of the events actually occurred.

How would the answer change if there were n balls and n boxes instead of 10 and 10?

The same method generalizes immediately. The probability that a specific box remains empty after n independent throws is ((n−1)/n)n, and by linearity of expectation, the expected number of empty boxes out of n boxes is n × ((n−1)/n)n, which simplifies to (n−1)n/nn−1. Setting n = 10 reproduces exactly the answer we derived here.

Is this the same idea as the "birthday problem"?

It's a close cousin. The birthday problem usually asks for the probability that at least two people share a birthday, whereas this question asks for the expected count of boxes that receive zero balls. Both problems live in the same "random allocation into cells" family and are commonly solved using similar probability setups, though the specific quantity being computed differs.

Could this expectation also be computed without indicator variables, by directly finding the probability distribution of N?

Yes, in principle you could find P(N = k) for each possible value of k from 0 to 9 using inclusion-exclusion counting arguments, and then compute E(N) as the weighted sum of k times P(N = k). However, this approach is far more tedious and error-prone than the indicator variable method, since it requires carefully accounting for overlapping cases of which specific boxes are empty. The indicator approach bypasses all of that complexity.

Why does the answer come out greater than 1 even though each individual box has less than a 50% chance of being empty?

Each box individually has probability (9/10)10 ≈ 0.349 of being empty, which is indeed less than one-half. But we are summing this probability across all 10 boxes, and 10 × 0.349 ≈ 3.487. There's nothing contradictory here: expectation of a count can exceed 1 even when each individual contributing probability is well below 1, as long as there are enough terms being added together.

If any step above felt unclear, or if you spotted a different way to approach this occupancy problem, drop a comment below — and if you found this walkthrough useful, do share it with a fellow ISS aspirant who might be revising the same topic.