Problem 71 - Calculus for Biology and Medicine

Question

An urn contains 5 green and 25 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

Expected value \( E(T) = 6 \); variance \( \operatorname{var}(T) = 30 \).

Step 1: Define the Probability of Drawing a Green Ball

There are 5 green balls and 25 blue balls in the urn, making a total of 30 balls. The probability of drawing a green ball, denoted as \( p \), is the ratio of green balls to the total number of balls: \( p = \frac{5}{30} = \frac{1}{6} \).

Step 2: Identify the Distribution of T

Since each draw is independent and the drawing continues until the first green ball is selected, the number of draws required, \( T \), follows a geometric distribution. The parameter of this distribution is \( p = \frac{1}{6} \).

Step 3: Calculate the Expected Value, E(T)

For a geometric distribution with success probability \( p \), the expected value, \( E(T) \), is given by \( E(T) = \frac{1}{p} \). Here, \( E(T) = \frac{1}{\frac{1}{6}} = 6 \).

Step 4: Calculate the Variance, Var(T)

The variance of a geometric distribution with parameter \( p \) is given by the formula \( \operatorname{var}(T) = \frac{1-p}{p^2} \). Here, \( \operatorname{var}(T) = \frac{1-\frac{1}{6}}{\left(\frac{1}{6}\right)^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = 30 \).

Key Concepts

Probability

Probability is a fundamental concept in statistics that measures the likelihood of an event occurring. In our example with the urn, we are interested in the chance of drawing a green ball. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

• There are 5 green balls out of a total of 30 balls, which include 25 blue balls.
• This makes the probability (p) of drawing a green ball equal to \(p = \frac{5}{30} = \frac{1}{6}\).

When each ball is replaced before drawing the next one, this scenario is known as sampling with replacement, ensuring that the probability remains constant throughout the process.

Expected Value

Expected value is essentially the average outcome you can expect if an experiment is repeated many times. In the context of a geometric distribution, like drawing balls with the condition of stopping at the first green ball, expected value gives us an insight into how many draws it might take on average before achieving a 'success' (drawing a green ball in this case).

• For a geometric distribution, the formula for expected value, denoted as \(E(T)\), is \(E(T) = \frac{1}{p}\), where \(p\) is the probability of success.
• In our example, since \(p = \frac{1}{6}\), we find that \(E(T) = \frac{1}{\frac{1}{6}} = 6\).

This indicates that on average, you would need to draw 6 balls to get a green ball. It's crucial to understand that expected value may not always be a number of draws you can achieve in reality, but rather a statistical average.

Variance

Variance provides a measure of how much the results are spread out from the expected value. It is especially useful when you want to understand the variability of your outcomes. In geometric distributions, where trials continue until the first success, variance gives us insight into the consistency of those trials.

• The formula for variance in a geometric distribution, represented as \(\operatorname{var}(T)\), is \(\operatorname{var}(T) = \frac{1-p}{p^2}\).
• In our problem, with \(p = \frac{1}{6}\), the variance calculates to \(\operatorname{var}(T) = \frac{1-\frac{1}{6}}{(\frac{1}{6})^2} = \frac{\frac{5}{6}}{\frac{1}{36}} = 30\).

The result indicates that while the average number of draws required is 6, the variance of 30 shows considerable spread around this mean. This means outcomes might fluctuate significantly around the expected value.