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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. What is the probability that the transferred ball was white given that the second ball drawn was white?

Short Answer

Expert verified
The probability that the transferred ball was white given that the second ball drawn was white is \(\frac{8}{17}\).

Step by step solution

01

Define the events of transferring and drawing balls

Let's define the following events: Event W1: Transferring a white ball from urn A to urn B. Event W2: Drawing a white ball from urn B after the transfer. We are finding the conditional probability P(W1|W2), which means the probability of transferring a white ball (W1) given that a white ball was drawn from urn B (W2).
02

Calculate the probability of transferring a white ball from urn A to urn B

Urn A has four white balls and six black balls. So, the probability of transferring a white ball (W1) is \( \frac{4}{(4+6)}\). P(W1) = \( \frac{4}{10}\) = \( \frac{2}{5}\)
03

Calculate the probability of drawing a white ball from urn B after the transfer

There are now two possibilities when drawing a white ball from urn B after the transfer: 1. If the transferred ball is white, urn B will have four white balls, and the probability of drawing a white ball (W2) is \( \frac{4}{9}\). 2. If the transferred ball is black, urn B will have three white balls, and the probability of drawing a white ball (W2) is \( \frac{3}{9}\).
04

Calculate the total probability of drawing a white ball from urn B

Now, we need to calculate the total probability of drawing a white ball from urn B. We can write this as: P(W2) = P(W2|W1) * P(W1) + P(W2|¬W1) * P(¬W1) Where ¬W1 is the event of not transferring a white ball, which means transferring a black ball instead. P(W2) = \( \left(\frac{4}{9}\right) * \left(\frac{2}{5}\right) + \left(\frac{3}{9}\right) *\left(\frac{3}{5}\right)\) = \( \frac{8}{45} + \frac{9}{45}\) = \( \frac{17}{45}\)
05

Calculate the conditional probability P(W1|W2)

Finally, we can find the conditional probability P(W1|W2) using the formula: P(W1|W2) = \(\frac{P(W1) * P(W2|W1)}{P(W2)}\) P(W1|W2) = \(\frac{\left(\frac{2}{5}\right) * \left(\frac{4}{9}\right)}{\frac{17}{45}}\) = \(\frac{8}{17}\) So, the probability that the transferred ball was white given that the second ball drawn was white is \(\frac{8}{17}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
At the core of many mathematical puzzles and real-world scenarios lies probability theory. This branch of mathematics is concerned with the likelihood of events occurring. It encompasses everything from the flip of a coin to more complex events like the problem involving the transfer of balls between urns A and B.

Understanding probability theory is crucial when dealing with conditional probability, such as in our textbook exercise. It involves calculating the probability of an event under the assumption that another event has already occurred. This framework of thinking helps in understanding the interconnectedness of events and how the outcome of one can influence the likelihood of another.

Key to grasp are terms like 'independent events', where the occurrence of one event does not affect the probability of another, and 'dependent events', where it does. Our exercise features dependent events - drawing a ball from urn A affects what's possible when picking from urn B, a fact that significantly alters the calculations of subsequent events and their probabilities.

Probability theory is also the foundation for various applications in fields as diverse as statistics, finance, gambling, science, and more.
Bayes' Theorem
Bayes' theorem is a powerful formula used to update the probability estimates of an event as more information becomes available. It plays a pivotal role in more advanced probability problems, such as the one we are examining with urns A and B. In essence, Bayes' theorem provides a way to reverse conditional probabilities.

When we are given that a white ball is drawn from urn B, Bayes' theorem allows us to recalculate the likelihood that this ball came from urn A. The formula hinges on the idea of prior probability, which is our initial assessment of an event's likelihood, and posterior probability, the updated chances after observing new evidence.

Our exercise illustrates Bayes' theorem in action. We start with an initial belief (the prior probability) about the presence of white balls and, upon observing the outcome of a white ball being drawn from urn B (our evidence), we revise our belief (to the posterior probability). This process shows how Bayes' theorem helps in reasoning backward from outcomes to causes, a fundamental method used in diverse fields such as diagnostic testing and machine learning.
Combinatorics
Combinatorics is the area of mathematics concerned with counting, both as a means to an end and as a pursuit in its own right. It deals with combinations and arrangements of objects according to specified rules. In probability theory, combinatorics is used to figure out the number of possible outcomes and hence calculate the probabilities of various events.

The exercise with urn A and B is an example of a combinatorial problem. Before transferring a ball from urn A to urn B, we must count the different ways in which a white or a black ball can be chosen. The complexity of the problem increases when the outcome of choosing from one urn affects the contents and the outcomes of choosing from the other.

Key concepts in combinatorics include permutations, combinations, and the Fundamental Principle of Counting. These tools allow one to calculate the probabilities of events without having to list every single outcome, which can be impractical or impossible with large sets of possibilities. Integrating combinatorial approaches streamlines the process of solving complex probability exercises, like the textbook problem we see with the two urns.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Let \(S=\left\\{s_{1}, s_{2}, \ldots, s_{n}\right\\}\) be a uniform sample space for an experiment. If \(n \geq 5\) and \(E=\left\\{s_{1}, s_{2}, s_{5}\right\\}\), then \(P(E)=3 / n\).

Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$\begin{array}{llllll} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array}$$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

Copykwik has four photocopy machines: \(A, B, C\), and \(D .\) The probability that a given machine will break down on a particular day is \(P(A)=\frac{1}{50} \quad P(B)=\frac{1}{60} \quad P(C)=\frac{1}{75} \quad P(D)=\frac{1}{40}\) Assuming independence, what is the probability on a particular day that a. All four machines will break down? b. None of the machines will break down?

According to the Centers for Disease Control and Prevention, the percentage of adults \(25 \mathrm{yr}\) and older who smoke, by educational level, is as follows: $$\begin{array}{lcccccc} \hline & & & \text { High } & {\text { Under- }} \\ \text { Educational } & \text { No } & \text { GED } & \text { school } & \text { Some } & \text { graduate } & \text { Graduate } \\ \text { Level } & \text { diploma } & \text { diploma } & \text { graduate } & \text { college } & \text { level } & \text { degree } \\ \hline \text { Respondents, \% } & 26 & 43 & 25 & 23 & 10.7 & 7 \\ \hline \end{array}$$ In a group of 140 people, there were 8 with no diploma, 14 with GED diplomas, 40 high school graduates, 24 with some college, 42 with an undergraduate degree, and 12 with a graduate degree. (Assume that these categories are mutually exclusive.) If a person selected at random from this group was a smoker, what is the probability that he or she is a person with a graduate degree?

Propuct Reuasiumr The proprietor of Cunningham's Hardware Store has decided to install floodlights on the premises as a measure against vandalism and theft. If the probability is \(.01\) that a certain brand of floodlight will burn out within a year, find the minimum number of floodlights that must be installed to ensure that the probability that at least one of them will remain functional for the whole year is at least .99999. (Assume that the floodlights operate independently.)

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