Problem 9 - A First Course in Probability

Question

A total of \(n\) balls, numbered 1 through \(n\), are put into \(n\) urns, also numbered 1 through \(n\) in such a way that ball \(i\) is equally likely to go into any of the urns \(1,2, \ldots, i .\) Find (a) the expected number of urns that are empty; (b) the probability that none of the urns is empty.

Step-by-Step Solution

Answer

In summary: (a) The expected number of urns that are empty is 1. (b) The probability that none of the urns is empty is given by the formula: \[ P(\text{no urn is empty}) = \frac{D_n}{n^n} = \frac{n!}{n^n}\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n\frac{1}{n!}\right) \]

Step 1: Understand the problem and define variables

We have \(n\) balls and \(n\) urns. Let \(E_i\) be the expected number of empty urns after the first \(i\) balls are thrown. We are interested in finding \(E_n\), the expected number of empty urns after all \(n\) balls are thrown.

Step 2: Calculate the probability of the ith urn being empty

To find the expected number of empty urns, we can consider the probability of each urn being empty at the end. We will calculate this probability for urn \(i\) and then sum the probabilities for all \(i\). The probability that urn \(i\) is empty after one ball has been thrown is \(1 - \frac{1}{i}\). The probability that urn \(i\) is empty after \(n\) balls have been thrown can be calculated using the product rule for independent events (since putting a ball into one urn doesn't affect the probabilities of putting balls into the other urns). So, the probability of urn \(i\) being empty after all \(n\) balls have been thrown is the product of the probabilities of urn \(i\) being empty after each ball has been thrown, which is \(P_i = \frac{1}{i} \times \frac{2}{i+1} \times \cdots \times \frac{n-1}{n} = \frac{1}{n}\).

Step 3: Calculate the expected number of empty urns

Now that we have the probability of urn \(i\) being empty, we can calculate the expected number of empty urns by taking the sum of the probabilities for all \(i\). Since the probability of each urn being empty is \(\frac{1}{n}\), we have \(E_n = \sum_{i=1}^n \frac{1}{n} = \frac{1}{n} \sum_{i=1}^n 1 = 1\). Hence, the expected number of empty urns is 1. (a) The expected number of urns that are empty is 1. Now, we need to find the probability that none of the urns is empty.

Step 4: Calculate the probability that no urns are empty using permutations

Let's find the number of ways of placing \(n\) balls into \(n\) urns such that no urn is empty. Notice that this is equivalent to finding the number of permutations of the balls in which no ball is in its corresponding urn. This is known as the derangement formula. Let \(D_n\) be the number of derangements of \(n\) elements, which can be calculated using the following formula, also called the inclusion-exclusion principle: \[D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n\frac{1}{n!}\right)\]

Step 5: Calculate the probability of no urn being empty

Now, we can calculate the probability of no urn being empty as the number of derangements divided by the total number of ways to distribute the balls in the urns, which is \(n^n\): \[ P(\text{no urn is empty}) = \frac{D_n}{n^n} = \frac{n!}{n^n}\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n\frac{1}{n!}\right) \] (b) This formula gives the probability that none of the urns is empty.

Key Concepts

Probability Theory

Probability theory is the mathematical framework used to analyze random events. In the context of the urn problem, it's about understanding how likely different outcomes are when balls are distributed into urns.

Here's how we broke it down:

• Each ball can go into any of a limited set of urns.
• The outcome of where one ball ends up doesn't influence where another ball can go.
• We consider each urn's chance of being empty after all balls are thrown your way.

The key realization is that each urn remains empty with a certain probability. We calculated these probabilities for all urns and found the average (or expected value) of empty urns. By summing these probabilities, we determine how many urns are likely to be empty. This aligns with the expected value of such random processes, where you weigh all possibilities by the chance they occur.

Inclusion-Exclusion Principle

The Inclusion-Exclusion Principle is a combinatorial method used to count the number of elements in the union of overlapping sets. It's handy for avoiding over-counting the elements that belong to multiple sets.

In our problem, this principle comes into play when calculating derangements, where we ensure no urn is left empty. Here's the breakdown:

• First, count all arrangements where balls could end up, considering any pattern.
• Subtract cases where at least one urn is empty.
• Keep correcting for overlaps of empty urns by alternating between adding and subtracting overlaps of different sizes.

Mathematically, we use this principle to derive the formula for derangements. The successive adding and subtracting mimic ensuring no urn is empty. This is crucial when calculating probabilities in setups with overlapping conditions.

Derangements

In combinatorics, a derangement is a permutation where no object appears in its original position. Applying this to the balls and urns, a derangement occurs when none of the balls land in its corresponding urn.

Why does this matter? To find out the probability that no urn remains empty, we need to ensure each urn has at least one ball. Derangements help us understand precisely how:

• Calculate the total permutations where no ball is in its numbered urn.
• Use the inclusion-exclusion principle to determine how often this happens.
• Represented by the formula: \[D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^n \frac{1}{n!}\right)\]

This formula gives the number of valid arrangements, ensuring every urn gets a ball. Derangements are vital for tasks requiring specific positioning constraints, like seating plans or this urn problem.