Problem 9 - A First Course in Probability

Question

Consider 3 urns. Urn \(A\) contains 2 white and 4 red balls, urn \(B\) contains 8 white and 4 red balls, and urn \(C\) contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn \(A\) was white given that exactly 2 white balls were selected?

Step-by-Step Solution

Answer

The probability that the ball chosen from urn A was white given that exactly 2 white balls were selected is \(\frac{7}{13}\).

Step 1: Identify the relevant events and probabilities

Let us define the events of interest: - Event A: A white ball is selected from urn A. - Event B: A white ball is selected from urn B. - Event C: A white ball is selected from urn C. - Event W: Exactly 2 white balls are selected in total. We are asked to find the probability of event A given event W, or \(P(A|W)\). First, we find the probabilities of selecting a white or red ball from each urn: - Urn A: \(P(A) = \frac{2}{6} = \frac{1}{3}\) and \(P(\overline{A}) = \frac{4}{6} = \frac{2}{3}\) - Urn B: \(P(B) = \frac{8}{12} = \frac{2}{3}\) and \(P(\overline{B}) = \frac{4}{12} = \frac{1}{3}\) - Urn C: \(P(C) = \frac{1}{4}\) and \(P(\overline{C}) = \frac{3}{4}\)

Step 2: Identify all combinations that can result in exactly 2 white balls

To find event W (exactly 2 white balls are selected), we check all possible combinations that satisfy this condition: 1. White from urn A, white from urn B, and red from urn C: \(A \cap B \cap \overline{C}\) 2. White from urn A, red from urn B, and white from urn C: \(A \cap \overline{B} \cap C\) 3. Red from urn A, white from urn B, and white from urn C: \(\overline{A} \cap B \cap C\)

Step 3: Calculate the probabilities of each combination in Step 2

Using the probabilities found in Step 1, we calculate the probabilities of the combinations in Step 2: 1. \(P(A \cap B \cap \overline{C}) = P(A) \times P(B) \times P(\overline{C}) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{6}\) 2. \(P(A \cap \overline{B} \cap C) = P(A) \times P(\overline{B}) \times P(C) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\) 3. \(P(\overline{A} \cap B \cap C) = P(\overline{A}) \times P(B) \times P(C) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{9}\) Now, we can find the probability of event W (exactly 2 white balls are selected): \(P(W) = P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C) + P(\overline{A} \cap B \cap C) = \frac{1}{6} + \frac{1}{36} + \frac{1}{9} = \frac{13}{36}\)

Step 4: Apply conditional probability formula and find the required probability

We now have everything we need to find the probability of event A given event W (a white ball is chosen from urn A given exactly 2 white balls are selected). Using the conditional probability formula: \(P(A|W) = \frac{P(A \cap W)}{P(W)} = \frac{P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C)}{P(W)}\) Plugging in the probabilities calculated in Steps 3 and 4: \(P(A|W) = \frac{\frac{1}{6} + \frac{1}{36}}{\frac{13}{36}} = \frac{7}{13}\) So the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected is \(\boxed{\frac{7}{13}}\).

Key Concepts

Probability Theory

Understanding probability theory is crucial for solving problems involving uncertainty and chance. It helps us quantify the likelihood of different outcomes. At its core, probability measures how likely something is to happen. For any given event, the probability will always be a number between 0 and 1. A probability of 0 means the event will not happen, while a probability of 1 means it will certainly happen.

When dealing with multiple events, it's important to know how to calculate the probability of various combinations. This exercise involves selecting balls from urns, where each selection is an event. Given different possibilities, we use probability theory to compute combinations such as two white balls out of three selections. You begin by defining events that represent the scenarios you're interested in. Then you compute their individual probabilities using basic probability rules. These include multiplying probabilities for independent events and summing probabilities for combined outcomes. This strategy allows us to calculate complex scenarios more systematically. Remember, clarity in defining events and methodically calculating probabilities can resolve many such problems efficiently.

Bayes' Theorem

Bayes' Theorem is a powerful tool in conditional probability that lets us update the probability of an event based on new information. It's especially useful when the situation involves prior knowledge or evidence that affects probability.

In this exercise, you are asked to find the probability that a white ball was chosen from urn A given a certain condition: specifically, that exactly two white balls were chosen overall. This is where Bayes' Theorem comes in handy, articulated as:

• \(P(A|B) = \frac{P(B|A) \times P(A)}{P(B)}\)

Here, \(P(A|B)\) is the probability of event A given that event B has occurred. The task is to determine this probability by considering all the possible ways event B could happen, then figuring out how likely event A is under those circumstances.

This exercise applies Bayes' Theorem by breaking it into steps, starting with calculating individual probabilities of key events (step-by-step). You then use these to find conditional probabilities, like \(P(A|W)\), where W is the event of selecting exactly two white balls. Understanding Bayes' Theorem can immensely simplify the process of dealing with conditional probabilities, especially in real-world applications.

Combinatorial Probability

Combinatorial probability combines the principles of probability with combinatorial counting techniques to address questions where multiple combinations and arrangements are possible. It’s all about determining how many specific ways events can occur, and what the probabilities of those occurrences are.

In this task, you're dealing with three urns, each containing different numbers of red and white balls. When you choose one ball from each urn, there are numerous possible outcomes, but only a subset of these outcomes include exactly two white balls. This scenario is where combinatorial methods become valuable, as they allow you to count these combinations effectively.

You have to consider all valid arrangements to ensure that exactly two white balls are selected. This involves determining three specific scenarios:

• White from urn A and B, red from urn C.
• White from urn A and C, red from urn B.
• Red from urn A, white from urn B and C.

For each case, compute its probability and sum these to get the total probability of exactly two white balls. By using combinations, you streamline the process of examining all possible outcomes, making complex probability problems more manageable.