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

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

Step 1: Define the Probability of Success

A green ball being drawn is considered a success. There are 10 green balls and 20 blue balls, making a total of 30 balls. The probability \(P\) of drawing a green ball (success) is \(P = \frac{10}{30} = \frac{1}{3}\).

Step 2: Determine the Type of Random Variable

The first time until an event occurs when trials are independent and the probability of success is constant in each trial follows a geometric distribution. Thus, \(T\) follows a geometric distribution with parameter \(p = \frac{1}{3}\).

Step 3: Find the Expectation of a Geometric Distribution

The expected value for a geometric distribution is \(E(T) = \frac{1}{p}\). Substituting \(p = \frac{1}{3}\), we find \[E(T) = \frac{1}{\frac{1}{3}} = 3.\]

Step 4: Find the Variance of a Geometric Distribution

The variance for a geometric distribution is \(\operatorname{var}(T) = \frac{1-p}{p^2}\). Substituting \(p = \frac{1}{3}\), we calculate \(\operatorname{var}(T)\) as follows: \[\operatorname{var}(T) = \frac{1 - \frac{1}{3}}{\left( \frac{1}{3} \right)^2} = \frac{\frac{2}{3}}{\frac{1}{9}} = 6.\]

Key Concepts

Probability Theory

Probability theory is an essential branch of mathematics that deals with the likelihood of an event occurring. It helps us understand how to assign probabilities to events and predict outcomes in uncertain scenarios. In our exercise with the urn containing green and blue balls, we focused on the probability of drawing a green ball. The probability of success, defined here as drawing a green ball, was calculated by dividing the number of green balls by the total number of balls.

• There are 10 green balls, giving us a probability of success: \( P = \frac{10}{30} = \frac{1}{3} \).

This demonstrates the core principle of probability theory, where outcomes are determined based on the ratio of favorable outcomes to total possible outcomes. Understanding this concept is crucial as it forms the basis for many probability distributions like the geometric distribution.
By knowing probabilities, we can predict and calculate further statistical properties such as the expected value and variance.

Expected Value

The expected value is a key concept in probability theory and statistics that gives us the long-term average of a random variable, considering all possible outcomes. In the context of a geometric distribution, such as our exercise where we first draw a green ball from an urn, the expected value tells us the average number of trials needed to achieve a success.

For a geometric distribution defined by a success probability \( p \), the expected value \( E(T) \) is calculated as \( E(T) = \frac{1}{p} \). In our exercise, with \( p = \frac{1}{3} \), the expected value calculates to 3. This means on average, we expect to draw three balls before obtaining a green ball.

This average helps us organize and predict outcomes over many trials, providing a foundation for understanding random processes.

Variance

Variance measures how much the outcomes of a random variable differ from the expected value and from each other. A higher variance indicates greater variability in the outcomes, while a lower variance suggests that outcomes are consistently closer to the average.

In a geometric distribution, the variance is determined by the formula \( \operatorname{var}(T) = \frac{1-p}{p^2} \), where \( p \) is the probability of success. Using our example, where \( p = \frac{1}{3} \), we find the variance to be 6.

• This means that there is moderate variability in the number of draws needed before a green ball is drawn.

Understanding variance allows us to grasp the spread of possible outcomes and gives us insight into the predictability of the process. It complements the expected value by showing us how much variation exists around this average.