Problem 71 - Calculus for Biology and Medicine

Question

An urn contains five 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

The expected time is 6 draws and the variance is 30 draws.

Step 1: Define the Geometric Distribution

In this problem, we need to determine both the expected value and the variance of the time until the first success (drawing a green ball). Here, drawing a green ball is considered a 'success'. Since each draw is independent and the conditions meet the properties of a geometric distribution, we use the geometric distribution framework.Let the probability of success (drawing a green ball) be denoted as \( p \). The probability of not drawing a green ball (failure) is \( q = 1 - p \).

Step 2: Calculate Probability of Success

To calculate the probability \( p \) of drawing a green ball from the urn, we need to know the ratio of green balls to the total number of balls in the urn. There are 5 green balls and 25 blue balls, giving a total of 30 balls.Therefore, the probability of drawing a green ball is:\[ p = \frac{5}{30} = \frac{1}{6} \]

Step 3: Expected Value of Geometric Distribution

For a geometric distribution with probability \( p \), the expected value \( E(T) \) is given by:\[ E(T) = \frac{1}{p} \]Substituting the value of \( p \) we found:\[ E(T) = \frac{1}{1/6} = 6 \]

Step 4: Variance of Geometric Distribution

The variance \( \operatorname{var}(T) \) of a geometric distribution with probability \( p \) is given by:\[ \operatorname{var}(T) = \frac{1-p}{p^2} \]We already found \( p = \frac{1}{6} \), so:\[ \operatorname{var}(T) = \frac{1 - 1/6}{(1/6)^2} = \frac{5/6}{1/36} = 30 \]

Step 5: Conclusion

We calculated the expected value and variance using properties of the geometric distribution. The expected number of draws until a green ball is drawn is 6, and the variance of the number of draws is 30.

Key Concepts

Probability

In probability theory, the concept of probability is the measure of the likelihood that an event will occur. When dealing with a geometric distribution, like in this exercise, probability helps us determine the success rate of obtaining a particular outcome in a series of trials.
A geometric distribution occurs when the same experiment is conducted repeatedly, with the results either being success or failure, until the first success is achieved. In this case, the experiment is drawing balls from an urn until a green ball (a success) is drawn.
To calculate the probability of the success, or probability of drawing a green ball, we use the formula: \[ p = \frac{\text{Number of Green Balls}}{\text{Total Number of Balls}} \]

• The number of green balls is 5 and the total number of balls is 30 (5 green + 25 blue).
• Therefore, the probability \( p \) of drawing a green ball is \( \frac{5}{30} = \frac{1}{6} \).

This probability \( p \) is crucial as it is used to calculate both the expected value and variance in a geometric distribution model.

Expected Value

The expected value is central in understanding the average outcome of a random process over time. In the context of geometric distribution, the expected value tells us the typical number of trials needed to achieve the first success.
For a geometric distribution, the expected value \( E(T) \) is determined by the reciprocal of the probability \( p \):\[ E(T) = \frac{1}{p} \]

• Here, \( p = \frac{1}{6} \).
• Thus, substituting the value into the formula gives \( E(T) = \frac{1}{1/6} = 6 \).

Therefore, on average, one would expect it to take 6 draws to find the first green ball. Knowing the expected value helps in predicting outcomes and making informed decisions in situations modeled by a geometric distribution.

Variance

Variance in probability provides insight into the spread of possible outcomes around the expected value. It measures how much the outcomes of a random variable deviate from what is "expected" over time.
In a geometric distribution, the variance \( \operatorname{var}(T) \) is calculated using the formula:\[ \operatorname{var}(T) = \frac{1-p}{p^2} \]

• Using \( p = \frac{1}{6} \), we find the variance as \( \operatorname{var}(T) = \frac{1 - 1/6}{(1/6)^2} = \frac{5/6}{1/36} = 30 \).

This indicates that the variance of the number of draws required to first draw a green ball is 30, which suggests that although on average it takes 6 draws, there can be a significant amount of deviation from this mean and a range of different outcomes might occur. Understanding the variance is beneficial in assessing the risk or uncertainty in making predictions based on the geometric distribution.