Problem 70 - Calculus for Biology and Medicine

Question

An urn contains one 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 is \(\left(\frac{n-1}{n}\right)^{n-1}\). As \(n \rightarrow \infty\), it approaches \(e^{-1}\).

Step 1: Understanding the Problem

We have an urn with one black ball and \(n-1\) white balls. We draw balls one by one until we get the black ball. The probability of drawing the black ball is \(\frac{1}{n}\) and drawing a white ball is \(\frac{n-1}{n}\). We want to find the probability that at least \(n\) draws are needed to get the black ball.

Step 2: Defining Probability of At Least n Draws

To have at least \(n\) draws, the black ball must not be drawn in the first \(n-1\) draws. This means that all \(n-1\) draws result in drawing a white ball, which has a probability of \(\frac{n-1}{n}\) each time.

Step 3: Calculating Probability of n-1 White Balls

The probability of drawing a white ball \(n-1\) times in succession is given by multiplying the probability of drawing a white ball each time: \(\left(\frac{n-1}{n}\right)^{n-1}\).

Step 4: State Final Probability Expression

Thus, the probability that at least \(n\) draws are needed to draw the black ball is \(\left(\frac{n-1}{n}\right)^{n-1}\).

Step 5: Exploring Limit as n Approaches Infinity

To understand what happens as \(n\) approaches infinity, we evaluate the limit: \[ \lim_{n \to \infty} \left(\frac{n-1}{n}\right)^{n-1} = e^{-1} \approx 0.3679 \]. This is because \(\left(\frac{n-1}{n}\right)^{n}\) approximates to \(e^{-1}\) when \(n\) tends to infinity.

Key Concepts

Limit of a sequence

A sequence is a collection of numbers ordered in a specific way, where each number in the series is the term of the sequence.
In probability theory, understanding the limit of a sequence helps us predict the behavior of probabilities as some variable approaches infinity.
For instance, in the original exercise, we look at the sequence generated by repeated draws from an urn. This sequence describes the probability of needing at least \( n \) draws to select a black ball. The mathematical expression for this is \( \left(\frac{n-1}{n}\right)^{n-1} \), which denotes the sequence.

• As \( n \) increases, the sequence progresses closer to a specific value, called the limit.
• The limit in this context tells us what happens with the probability as the number of draws becomes very large.

Analyzing this limit, \( \lim_{n \to \infty} \left(\frac{n-1}{n}\right)^{n-1} = e^{-1} \), gives us the surprising result that the probability stabilizes around 0.3679. Even as the number of possible draws increases indefinitely, the likelihood doesn't vanish or explode but settles at a constant percentage.

Infinite series

Infinite series add an important dimension to sequences, where instead of looking at the behavior of a single series of probabilities, one might explore an infinite sum of different terms derived from a sequence.
In probability, infinite series are often used to determine the total probability over all possible events.
A classic example includes geometric series seen in scenarios similar to our urn exercise.

• The geometric series relates to situations where events can happen repeatedly, each time with a fixed probability.
• Sum of an infinite geometric series is instrumental in expressing probabilities over an unlimited number of attempts.

While not directly applying in the original exercise since we deal more with sequences than series, the idea is central to understanding how repetitive processes behave when extended to infinity. In nature, these concepts help in calculating probabilities for ongoing events like rolling a dice until getting a certain number or repeating draws from an urn systemically. These processes naturally segue into discussions on expected values, particularly when the number of required trials isn't capped.

Expected value

Expected value offers a powerful way to average out possibilities in probabilistic events, providing a long-term mean behavior of random variables.
It's like the theoretical mean of infinite trials of a random event.
In our exercise, while we calculated probability, expected value would concern predicting on average, how many draws might be necessary to obtain a black ball.

• Calculated as the sum of all possible values of a random variable, each weighted by its probability.
• Provides insight on what might happen "on average" in a probabilistic setting.

In games of chance or predicting outcomes over time, expected value becomes crucial. It's determined by finding the most likely position of drawn events if you repeatedly execute the condition (e.g., pulling balls from an urn).
Use of expected value gives an optimal outcome measure, especially in contexts requiring decision-making, where understanding the likely number of attempts is beneficial. It steers decisions with a numerical representation of average occurrence, therefore driving better predictions. This forms an integral foundation in both theoretical and applied probability.