Problem 40 - A First Course in Probability

Question

Consider a sample of size 3 drawn in the following manner: We start with an urn containing 5 white and 7 red balls. At each stage, a ball is drawn and its color is noted. The ball is then returned to the urn, along with an additional ball of the same color. Find the probability that the sample will contain exactly. (a) 0 white balls; (b) 1 white ball; (c) 3 white balls; (d) 2 white balls.

Step-by-Step Solution

Answer

The probabilities for drawing each number of white balls in a sample of size 3 are: (a) 0 White Balls: approximately 0.2308 or \(\frac{252}{1092}\) (b) 1 White Ball: approximately 0.3206 or \(\frac{350}{1092}\) (c) 3 White Balls: approximately 0.0962 or \(\frac{105}{1092}\) (d) 2 White Balls: approximately 0.3522 or \(\frac{385}{1092}\)

Step 1: (a) 0 White Balls (3 Red Balls)#

To find the probability of drawing 3 red balls, we need to determine the probability at each stage: 1. Probability of drawing a red ball on the first draw: \(P(R1) = \frac{7}{12}\) 2. After returning a red ball, there will be 13 balls in the urn (8 red and 5 white). Probability of drawing a red ball on the second draw: \(P(R2|R1) = \frac{8}{13}\) 3. After returning a red ball, there will be 14 balls in the urn (9 red and 5 white). Probability of drawing a red ball on the third draw: \(P(R3|R2) = \frac{9}{14}\) Now, we can multiply all these probabilities, as they are conditional: \(P(0 \, White \, Balls) = P(R1) \cdot P(R2|R1) \cdot P(R3|R2) = \frac{7}{12} \cdot \frac{8}{13} \cdot \frac{9}{14} = \frac{252}{1092} \approx 0.2308\)

Step 2: (b) 1 White Ball (2 Red Balls)#

To find the probability of drawing 1 white and 2 red balls, we need to determine the probability at each stage in different orders (WRW, WRR, RWR, RWRR, RRW): - Order WRW: 1. Probability of drawing a white ball on the first draw: \(P(W1) = \frac{5}{12}\) 2. After returning a white ball, there will be 13 balls in the urn (7 red and 6 white). Probability of drawing a red ball on the second draw: \(P(R2|W1) = \frac{7}{13}\) 3. After returning a red ball, there will be 14 balls in the urn (8 red and 6 white). Probability of drawing a red ball on the third draw: \(P(R3|R2) = \frac{8}{14}\) - Order WRR: 1. Probability of drawing a white ball on the first draw: \(P(W1) = \frac{5}{12}\) 2. After returning a white ball, there will be 13 balls in the urn (7 red and 6 white). Probability of drawing a red ball on the second draw: \(P(R2|W1) = \frac{7}{13}\) 3. After returning a red ball, there will be 14 balls in the urn (8 red and 6 white). Probability of drawing a white ball on the third draw: \(P(W3|R2) = \frac{6}{14}\) - Order RWR: 1. Probability of drawing a red ball on the first draw: \(P(R1) = \frac{7}{12}\) 2. After returning a red ball, there will be 13 balls in the urn (8 red and 5 white). Probability of drawing a white ball on the second draw: \(P(W2|R1) = \frac{5}{13}\) 3. After returning a white ball, there will be 14 balls in the urn (8 red and 6 white). Probability of drawing a red ball on the third draw: \(P(R3|W2) = \frac{8}{14}\) Now, we can multiply all these probabilities and add up results for all orders: \(P(1 \, White \, Ball) = P(W1) \cdot P(R2|W1) \cdot P(R3|R2) + P(W1) \cdot P(R2|W1) \cdot P(W3|R2) \\ + P(R1) \cdot P(W2|R1) \cdot P(R3|W2) = \frac{5}{12} \cdot \frac{7}{13} \cdot \frac{8}{14} + \frac{5}{12} \cdot \frac{7}{13} \cdot \frac{6}{14} \\ + \frac{7}{12} \cdot \frac{5}{13} \cdot \frac{8}{14} = \frac{350}{1092} \approx 0.3206\)

Step 3: (c) 3 White Balls (0 Red Balls)#

To find the probability of drawing 3 white balls, we need to determine the probability at each stage: 1. Probability of drawing a white ball on the first draw: \(P(W1) = \frac{5}{12}\) 2. After returning a white ball, there will be 13 balls in the urn (7 red and 6 white). Probability of drawing a white ball on the second draw: \(P(W2|W1) = \frac{6}{13}\) 3. After returning a white ball, there will be 14 balls in the urn (7 red and 7 white). Probability of drawing a white ball on the third draw: \(P(W3|W2) = \frac{7}{14}\) Now, we can multiply all these probabilities, as they are conditional: \(P(3 \, White \, Balls) = P(W1) \cdot P(W2|W1) \cdot P(W3|W2) = \frac{5}{12} \cdot \frac{6}{13} \cdot \frac{7}{14} = \frac{105}{1092} \approx 0.0962\)

Step 4: (d) 2 White Balls (1 Red Ball)#

Since the probability of all possibilities sums up to 1, if we subtract the probabilities of the other three cases from the total, we will find the probability of having 2 white balls and 1 red ball: \(P(2 \, White \, Balls) = 1 - P(0 \, White \, Balls) - P(1 \, White \, Ball) - P(3 \, White \, Balls) \\ = 1 - \frac{252}{1092} - \frac{350}{1092} - \frac{105}{1092} = \frac{385}{1092} \approx 0.3522\) In conclusion, the probabilities for drawing each number of white balls in a sample of size 3 are: (a) 0 White Balls: approximately 0.2308 or \(\frac{252}{1092}\) (b) 1 White Ball: approximately 0.3206 or \(\frac{350}{1092}\) (c) 3 White Balls: approximately 0.0962 or \(\frac{105}{1092}\) (d) 2 White Balls: approximately 0.3522 or \(\frac{385}{1092}\)

Key Concepts

Conditional Probability

Conditional probability helps us understand the likelihood of an event occurring given that another event has already happened. In the exercise, each draw depends on the outcome of the previous one, showcasing a classic scenario where conditional probability is applied.
For example, after the first red ball is drawn, the probability of drawing a red ball on the second draw is influenced by the first outcome.
This is expressed as \( P(R2|R1) \), which means "the probability of drawing a red ball on the second draw given a red ball was drawn on the first." This formula captures the essence of conditional probability, allowing us to better predict outcomes in dependent events.

Sampling with Replacement

Sampling with replacement is a method used where each item drawn from a collection is returned before the next draw. In this exercise, it implies after picking a ball, we not only put it back but also add an identical one to the urn.
This way, the composition of the urn changes after each draw, affecting subsequent probabilities.
This method maintains the independence of events in terms of available choices but alters the total outcomes, reflecting a dynamic probability calculation across the draws. This mechanism ensures that each draw slightly modifies the original set-up, creating new probabilities for each cycle.

Probability Distribution

Probability distribution is a concept that helps us describe all possible outcomes of a random experiment and their corresponding probabilities. In the context of this problem, it refers to how the various numbers of white balls (0, 1, 2, or 3) in the sample are distributed.
The result includes quantifying the likelihood of each scenario, like having all red or all white balls, creating a distribution of probable outcomes for different compositions of the sample.
It's crucial to keep in mind that distributions like these serve to represent the chances of different scenarios, helping to visualize and calculate real-world probabilities effectively.

Mathematical Expectation

Mathematical expectation, often known as expected value, describes what we would expect on average from a probabilistic event. It provides a measure of the center or "average" of a probability distribution.
Regarding this problem, it informs us about the average number of white balls we might expect in a sample of size 3.
The formula for mathematical expectation is \( E(X) = \sum x \, P(x) \), where \( x \) represents the outcomes (0, 1, 2, or 3 white balls), and \( P(x) \) their probabilities. By calculating the expected values, students can understand the average behavior of such random experiments over many repetitions.