Problem 72 - Calculus for Biology and Medicine

Question

An urn contains 10 green and 20 blue balls. Balls are drawn at random, one at a time, until a green ball is selected. Each ball is replaced before the next ball is drawn. Let \(T\) denote the first time until a green ball is drawn. Find \(E(T)\) and \(\operatorname{var}(T)\).

Step-by-Step Solution

Answer

\(E(T) = 3\), \(\operatorname{var}(T) = 6\).

Step 1: Understanding the Problem

We have an urn with 10 green and 20 blue balls, making a total of 30 balls. Balls are drawn with replacement, and we are interested in determining the expected value and variance of the first draw that results in a green ball, denoted as \(T\).

Step 2: Define the Probability

The probability of drawing a green ball from the urn is \(P(G) = \frac{10}{30} = \frac{1}{3}\), since there are 10 green balls out of a total of 30 balls.

Step 3: Identify the Distribution

Since each draw is independent and we continue until we draw a green ball, \(T\) follows a geometric distribution with success probability \(p = \frac{1}{3}\).

Step 4: Calculate the Expectation

For a geometrically distributed random variable \(T\) with success probability \(p\), the expected value \(E(T)\) is given by \(\frac{1}{p}\). Here, \(E(T) = \frac{1}{\frac{1}{3}} = 3\).

Step 5: Calculate the Variance

The variance of a geometric random variable \(T\) with success probability \(p\) is given by \(\operatorname{var}(T) = \frac{1-p}{p^2}\). By substituting \(p = \frac{1}{3}\), we find \(\operatorname{var}(T) = \frac{1-\frac{1}{3}}{(\frac{1}{3})^2} = \frac{\frac{2}{3}}{\frac{1}{9}} = 6\).

Key Concepts

Expected Value

The concept of expected value, often referred to as the mean, represents the average outcome if a random process is repeated many times. In the context of a geometric distribution, the expected value indicates the average number of trials needed to get the first success. For our urn example, each draw is a trial, and drawing a green ball is considered a success. Since the probability of success \( p \) (drawing a green ball) is \( \frac{1}{3} \), the expected value \( E(T) \) for the first success is calculated as \( \frac{1}{p} \). This yields \( E(T) = \frac{1}{\frac{1}{3}} = 3 \). This means, on average, we expect to draw three balls before a green one appears.

• The expected value is calculated as \( \frac{1}{p} \) for geometric distributions.
• This indicates the average number of trials for the first success.
• In our problem, \( p = \frac{1}{3} \), leading to \( E(T) = 3 \).

Variance

Variance measures the spread of a random variable's possible outcomes and gives insight into the variability of these outcomes. In simpler terms, it tells us how much the number of trials can deviate from the expected value. For a geometric distribution, the formula for variance is \( \operatorname{var}(T) = \frac{1-p}{p^2} \). Using our urn problem with \( p = \frac{1}{3} \), we compute the variance as \( \frac{1 - \frac{1}{3}}{\left(\frac{1}{3}\right)^2} = 6 \). This outcome indicates quite a significant variance, suggesting that while we generally expect a green ball in three trials, the actual number can vary widely among repetitions of the experiment.

• Variance for geometric distribution: \( \frac{1-p}{p^2} \).
• Variance indicates how much the results fluctuate around the mean.
• In our exercise, variance is \( 6 \), showing a large spread in outcomes.

Probability of Success

In probability theory, the probability of success \( p \) is a fundamental concept, especially when dealing with stochastic processes like in our exercise. For a geometric distribution, each trial is independent, and each outcome can either result in success (drawing a green ball) or failure (drawing a blue ball). In our urn problem, the probability of drawing a green ball from a total of 30 balls (10 green and 20 blue) is calculated as \( \frac{10}{30} = \frac{1}{3} \). Each draw is an independent event since the ball is replaced after each draw. This constant probability \( p \) is crucial as it remains steady throughout the drawing process. Thus, for each draw, the chance remains \( \frac{1}{3} \) that it will result in a green ball.

• Probability of success \( p = \frac{number \ of \ successes}{total \ number \ of \ trials} \).
• In the urn exercise, \( p = \frac{1}{3} \) as there are 10 green balls out of 30.
• This probability is used to calculate both the expected value and variance.