Math Problem Statement

Let k and n be positive integers with k ≤ n. We have an urn containing n balls labeled 1, 2, . . . , n, and draw k balls without replacement uniformly at random. Find the expected value of the label of the smallest ball.

Solution

We are asked to find the expected value of the label of the smallest ball drawn when kk balls are drawn without replacement from an urn containing nn balls labeled 1,2,,n1, 2, \ldots, n. Let's work through the solution step-by-step.


1. Problem Setup

  • There are nn labeled balls: {1,2,,n}\{1, 2, \ldots, n\}.
  • We draw kk balls without replacement.
  • We are interested in the expected value of the smallest label among the kk chosen balls.

2. Key Insight: Distribution of the Smallest Ball

Let XX denote the label of the smallest ball drawn. Since the balls are drawn uniformly at random, each subset of size kk is equally likely. We aim to find the expected value of XX, which can be computed as:

E[X]=x=1nxPr(X=x)\mathbb{E}[X] = \sum_{x=1}^{n} x \cdot \Pr(X = x)


3. Computing the Probability Pr(X=x)\Pr(X = x)

For the smallest ball to have label X=xX = x, all other k1k - 1 balls chosen must have labels from the set {x+1,,n}\{x + 1, \ldots, n\}. The number of ways to choose k1k - 1 balls from this set is:

(nxk1)\binom{n - x}{k - 1}

Since there are (nk)\binom{n}{k} ways to choose any kk balls from the nn balls, the probability that the smallest ball has label xx is:

Pr(X=x)=(nxk1)(nk)\Pr(X = x) = \frac{\binom{n - x}{k - 1}}{\binom{n}{k}}


4. Setting Up the Expected Value Formula

We now substitute the expression for Pr(X=x)\Pr(X = x) into the expected value formula:

E[X]=x=1nx(nxk1)(nk)\mathbb{E}[X] = \sum_{x=1}^{n} x \cdot \frac{\binom{n - x}{k - 1}}{\binom{n}{k}}

We can factor out the constant 1(nk)\frac{1}{\binom{n}{k}}:

E[X]=1(nk)x=1nx(nxk1)\mathbb{E}[X] = \frac{1}{\binom{n}{k}} \sum_{x=1}^{n} x \cdot \binom{n - x}{k - 1}


5. Simplifying the Sum

This sum can be evaluated using known combinatorial identities, but the key idea is that the final result is known to simplify to:

E[X]=n+1k+1\mathbb{E}[X] = \frac{n + 1}{k + 1}


6. Final Answer

The expected value of the label of the smallest ball is:

n+1k+1\boxed{\frac{n + 1}{k + 1}}


Summary

The intuition behind the result is that the labels are evenly spread among the nn balls, and when drawing kk balls, the expected value of the smallest label corresponds to this balanced formula.


Would you like further clarifications or more details on any step?
Here are some related questions to explore:

  1. How would the result change if the draws were made with replacement?
  2. What is the variance of the smallest ball's label?
  3. How would the expected value change if we were interested in the largest label drawn?
  4. Can we generalize this result to continuous distributions (e.g., picking random real numbers)?
  5. How does the distribution of the smallest ball's label behave for large nn and small kk?

Tip: Recognizing patterns in combinatorial sums is often key to finding elegant solutions to complex probability problems.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Combinatorics
Expected Value

Formulas

E[X] = (n + 1) / (k + 1)
Pr(X = x) = binom(n - x, k - 1) / binom(n, k)

Theorems

Combinatorial Probability
Linear Expectation

Suitable Grade Level

College Level (Undergraduate Probability/Combinatorics)