Problem 70 - Calculus for Biology and Medicine

Question

An urn contains 1 black and \(n-1\) white balls. Balls are drawn at random, one at a time, until the black ball is selected. Each ball is replaced before the next ball is drawn. Find the probability that at least \(n\) draws are needed. What happens as \(n \rightarrow \infty ?\)

Step-by-Step Solution

Answer

The probability approaches \(1/e\) as \(n\to\infty\).

Step 1: Understand the Problem

We are given an urn with 1 black ball and \(n-1\) white balls. Our task is to find the probability that at least \(n\) draws are required to select the black ball from this urn, with replacement of each drawn ball.

Step 2: Define the Probability of Each Draw

Since there are \(n\) balls in total (1 black and \(n-1\) white), the probability of drawing the black ball in a single draw is \(\frac{1}{n}\), and the probability of drawing a white ball is \(\frac{n-1}{n}\).

Step 3: Calculate Probability of at Least \(n\) Draws

To require at least \(n\) draws, the black ball must not be drawn in the first \(n-1\) draws. The probability of drawing a white ball each time for \(n-1\) successive draws is given by \(\left( \frac{n-1}{n} \right)^{n-1}\).

Step 4: Consider the Outcome as \(n\rightarrow\infty\)

As \(n\) increases, the expression \(\left( \frac{n-1}{n} \right)^{n}\) approaches \(e^{-1}\). However, since we consider \(n-1\) draws here, as \(n\rightarrow\infty\), the probability \(\left( \frac{n-1}{n} \right)^{n-1}\) also approaches \(e^{-1}\), therefore decreasing.

Key Concepts

Understanding the Urn Problem

The urn problem is a classic framework used in probability theory to model random experiments involving objects. In this problem, an urn contains a certain number of items (often colored balls) which are drawn randomly. Here, we have an urn with one black ball and \(n-1\) white balls. Balls are drawn one at a time, and each is replaced before the next draw. This replacement keeps the total number of balls constant, altering the outcome probabilities in subsequent draws. Understanding how these probabilities are structured is key to solving our problem: finding the probability that selecting the black ball requires at least \(n\) draws.
For each draw, the set-up remains the same because of replacement, making it a "with-replacement" problem. This condition leads to independent draws. Concepts like independence and replacement are crucial here, as they define how likely specific outcomes are based on past draws (or lack thereof).

Executing Probability Calculation

Probability calculation helps us quantify the likelihood of an event, given a context like our urn problem. To evaluate the probability of needing at least \(n\) draws to select the black ball, it's imperative to look at the individual and cumulative probabilities of each event.

• The probability of drawing the black ball in a single draw is \( \frac{1}{n} \), given there are \(n\) balls total, one of which is black.
• The probability of drawing a white ball, which is more probable in this case, is \( \frac{n-1}{n} \).

This insights leads us to the likelihood of not drawing a black ball in the first \(n-1\) attempts, calculated as \( \left( \frac{n-1}{n} \right)^{n-1} \). This exponential calculation comes from multiplying the probability of drawing a white ball (\( \frac{n-1}{n} \)) by itself \(n-1\) times, covering each of the initial \(n-1\) draws.

Analyzing the Limit as n Approaches Infinity

When contemplating what happens as the number of balls \(n\) increases indefinitely, we enter into the realm of limits, a fundamental concept in calculus and probability. This involves observing how expressions and probabilities behave as values grow larger.
As \(n\rightarrow\infty\), the expression \( \left( \frac{n-1}{n} \right)^{n-1} \) is key. Intuitively, as \(n\) increases, \( \frac{n-1}{n} \) gets closer to 1, but it is slightly less than 1 due to being divided by a number just slightly larger. When raised to a large power like \(n-1\), the value of this expression trends towards \(e^{-1}\) or roughly 0.3679, meaning it gets smaller as \(n\) becomes very large.
This behavior illustrates how a probability may approach a limiting value, helping us understand the diminishing likelihood of drawing only white balls as the number of chances and the pool of balls increase.