Problem 15 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Refer to the following experiment: Urn A contains four white and six black balls. Urn B contains three white and five black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. Represent the probabilities associated with this two-stage experiment in the form of a tree diagram.

Step-by-Step Solution

Answer

The tree diagram for this two-stage experiment is as follows: /--- (W, \(P(W_A)= \frac{4}{10}\)) ----- (W, \(P(W_B|W_A) = \frac{4}{9}\)) / Urn A ----- (B, \(P(B_A) = \frac{6}{10}\)) ----- (W, \(P(W_B|B_A)= \frac{3}{9}\)) \ \--- (W, \(P(W_A)= \frac{4}{10}\)) ----- (B, \(P(B_B|W_A) = \frac{5}{9}\)) /--- (B, \(P(B_A) = \frac{6}{10}\)) ----- (B, \(P(B_B|B_A) = \frac{6}{9}\)) / Urn A

Step 1: Identify possible outcomes

To create a tree diagram, we first need to identify all possible outcomes for each stage in the experiment. We draw a ball from Urn A and transfer it to Urn B. There are two possible outcomes at this stage, drawing a white ball (W) or a black ball (B). Then we draw a ball from Urn B, which again can have two possible outcomes, drawing a white ball (W) or a black ball (B).

Step 2: Calculate the probabilities for each possible outcome

For this step, we will calculate the probabilities of drawing a white or black ball from Urn A and transferring it to Urn B, as well as drawing a white or black ball from Urn B. - Probabilities for Urn A: \(P(W_A) = \frac{4}{10}\) (the probability of drawing a white ball from Urn A) \(P(B_A) = \frac{6}{10}\) (the probability of drawing a black ball from Urn A) - Probabilities for Urn B: If we transfer a white ball to Urn B, there will be four white balls and five black balls in total. The probabilities then become: \(P(W_B|W_A) = \frac{4}{9}\) (the probability of drawing a white ball from Urn B given that we have transferred a white ball) \(P(B_B|W_A) = \frac{5}{9}\) (the probability of drawing a black ball from Urn B given that we have transferred a white ball) If we transfer a black ball to Urn B, there will be three white balls and six black balls in total. The probabilities then become: \(P(W_B|B_A)= \frac{3}{9}\) (the probability of drawing a white ball from Urn B given that we have transferred a black ball) \(P(B_B|B_A) = \frac{6}{9}\) (the probability of drawing a black ball from Urn B given that we have transferred a black ball)

Step 3: Represent the probabilities in a tree diagram

Now that we have calculated the probabilities, we will create a tree diagram representing the two-stage experiment. 1. Draw the first set of branches representing the ball drawn from Urn A: one branch for a white ball (W) and another branch for a black ball (B). Label the branches with their respective probabilities (i.e., \(P(W_A) = \frac{4}{10}\) and \(P(B_A) = \frac{6}{10}\)). 2. For each branch in the first set, draw two new branches representing the ball drawn from Urn B: one branch for a white ball (W) and another branch for a black ball (B). Label the branches with their respective probabilities (i.e., \(P(W_B|W_A) = \frac{4}{9}\), \(P(B_B|W_A) = \frac{5}{9}\), \(P(W_B|B_A)= \frac{3}{9}\), and \(P(B_B|B_A) = \frac{6}{9}\)). The tree diagram would look like this: /--- (W, \(P(W_A)= \frac{4}{10}\)) ----- (W, \(P(W_B|W_A) = \frac{4}{9}\)) / Urn A ----- (B, \(P(B_A) = \frac{6}{10}\)) ----- (W, \(P(W_B|B_A)= \frac{3}{9}\)) \ \--- (W, \(P(W_A)= \frac{4}{10}\)) ----- (B, \(P(B_B|W_A) = \frac{5}{9}\)) /--- (B, \(P(B_A) = \frac{6}{10}\)) ----- (B, \(P(B_B|B_A) = \frac{6}{9}\)) / Urn A

Key Concepts

Conditional Probability

Conditional probability is fundamental to understanding the likelihood of an event given that another event has already occurred. This concept is crucial when we deal with sequential experiments, where the outcome of one affects the probabilities in the next.

For instance, in the problem with the urns, after transferring a white ball from Urn A to Urn B, the composition of Urn B changes, which in turn alters the probability of drawing a particular color ball from Urn B. The notation, such as \(P(W_B|W_A)\), signifies the probability of drawing a white ball from Urn B given that a white ball was added to it from Urn A. Similarly, \(P(B_B|W_A)\) represents the probability of drawing a black ball from Urn B after a white ball was transferred. Here, ‘|’ symbolizes 'given that'.

Understanding conditional probability is crucial for correct probability calculation as it sets the stage for a more realistic assessment of events in a multi-stage setting, where each event can condition the next.

Two-Stage Experiment

A two-stage experiment involves two sequential events where the outcome of the first impacts the possible outcomes of the second. Our urn example perfectly illustrates this concept: drawing from Urn A is the first stage, and drawing from Urn B is the second stage, with the added twist that the result of the first draw actually changes the composition of Urn B.

When creating a probability tree diagram, it reflects the progressive nature of the experiment. The branches at each stage correspond to the possible outcomes at that point in time. We start by calculating the probabilities for the initial stage, and then adjust the probabilities for the outcomes of subsequent stages based on the previous results. This two-stage experiment approach can be extended to more complicated multi-stage processes, which are found in a multitude of real-world scenarios from genetics to finance.

Probability Calculation

Probability calculation involves determining the chance of a particular event occurring. It can range from very simple scenarios, like flipping a coin, to more complex, multi-layered situations, like our urn experiment.

Mathematically, probability is expressed as a fraction or a decimal between 0 and 1, where 0 indicates an impossible event and 1 denotes a certainty. In the context of our urns, the calculations start with the basic probabilities of drawing a white or black ball from Urn A, \(P(W_A)\) and \(P(B_A)\), respectively. Moving to Urn B, we then calculate conditional probabilities, adjusting the total number of balls in the urn after the transfer from Urn A. These steps are essential in systematically uncovering the likelihood of each possible outcome in the tree diagram.

Clarity in probability calculation is essential, as misunderstandings can lead to significant errors. Offering step-by-step solutions helps ensure that students can follow the logic behind each calculation, reducing confusion and bolstering their conceptual understanding.