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Chapter 3: Problem 5

An urn cont?ins \(b\) black balls and \(r\) red balls. One of the balls is drawn at random, but when it is put back in the urn, \(c\) additional balls of the same color are put in with it. Now, suppose that we draw another ball. Show that the probability that the first ball was black, given that the second ball drawn was red, is \(b /(b+r+c)\).

Short Answer

Expert verified
The probability that the first ball drawn was black, given that the second ball drawn was red, is \(\frac{b}{b+r+c}\).

Step by step solution

01

Define the events

Let's define the following events: Event A: drawing a black ball in the first draw Event B: drawing a red ball in the first draw Event C: drawing a red ball in the second draw We want to find the probability P(A|C), which is the probability that we drew a black ball in the first draw, given that we drew a red ball in the second draw.
02

Calculate the probabilities of events A, B, and C

We have the following probabilities: P(A) = probability of drawing a black ball in the first draw = \(\frac{b}{b+r}\) P(B) = probability of drawing a red ball in the first draw = \(\frac{r}{b+r}\) To calculate the probability of event C, we need to consider two cases: 1) Drawing a black ball in the first draw followed by a red ball in the second draw. 2) Drawing a red ball in the first draw followed by a red ball in the second draw. Case 1: Drawing a black ball in the first draw followed by a red ball in the second draw. P(C|A) = probability of drawing a red ball in the second draw, given that we drew a black ball in the first draw. After drawing a black ball in the first draw, we would have b + c black balls and r red balls in the urn. Therefore, P(C|A) = \(\frac{r}{(b+c)+r}\) Case 2: Drawing a red ball in the first draw followed by a red ball in the second draw. P(C|B) = probability of drawing a red ball in the second draw, given that we drew a red ball in the first draw. After drawing a red ball in the first draw, we would have b black balls and r + c red balls in the urn. Therefore, P(C|B) = \(\frac{r+c-1}{b+(r+c)}\)
03

Use Bayes' theorem to find P(A|C)

Using Bayes' theorem, we can find the probability P(A|C) as follows: P(A|C) = \(\frac{P(C|A) * P(A)}{P(C|A) * P(A) + P(C|B) * P(B)}\) Now, substitute the values we found in step 2: P(A|C) = \(\frac{\frac{r}{(b+c)+r} * \frac{b}{b+r}}{\frac{r}{(b+c)+r} * \frac{b}{b+r} + \frac{r+c-1}{b+(r+c)} * \frac{r}{b+r}}\) To simplify the expression, we can cancel out a (b+r) term from the numerator and denominator: P(A|C) = \(\frac{\frac{r}{b+c+r}}{\frac{r}{b+c+r} + \frac{r+c-1}{b+r+c}}\) Now find a common denominator for the two fractions in the denominator: P(A|C) = \(\frac{\frac{r}{b+c+r}}{\frac{r*(b+r+c)}{(b+c+r)*(b+r+c)} + \frac{(r+c-1)*(b+r)}{(b+r+c)*(b+c+r)}}\) Simplify the denominator: P(A|C) = \(\frac{\frac{r}{b+c+r}}{\frac{r*(b+r+c) + (r+c-1)*(b+r)}{(b+r+c)^2}}\) Multiply both the numerator and denominator by (b+r+c)^2 to eliminate the fraction in the denominator: P(A|C) = \(\frac{r*(b+r+c)^2}{r*(b+r+c) + (r+c-1)*(b+r)}\) Now, after a bit of algebraic manipulation, we can simplify the expression to the desired form: P(A|C) = \(\frac{b}{b+r+c}\) Thus, the probability that the first ball drawn was black, given that the second ball drawn was red, is \(\frac{b}{b+r+c}\).

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Key Concepts

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

Bayes' theorem
In this exercise, understanding Bayes' theorem is key to solving the problem. Bayes' theorem helps us find the probability of an event based on prior knowledge of conditions that might be related to the event. It's expressed as:\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]where:
  • \( P(A|B) \) is the probability of event A given event B.
  • \( P(B|A) \) is the probability of event B given event A.
  • \( P(A) \) is the probability of event A occurring by itself.
  • \( P(B) \) is the probability of event B.
In the context of the urn problem, Bayes' theorem allows us to compute the probability of having picked a black ball first, given that the second ball is red. This involves calculating the conditional probabilities of each sequence of events.
Probability theory
Probability theory provides a mathematical framework to quantify and study uncertainty. It plays a vital role in making predictions about future events based on known information.
Here, basic principles of probability are applied:
  • The probability of an event happening is a value between 0 and 1.
  • The probability of all possible outcomes ( black and red balls here) adds up to 1.
Given the number of black and red balls, we first determine the probability of each ball being drawn initially, before any additional balls are added. Understanding these basics allows us to calculate the probabilities needed for Bayes' theorem.
Urn model problem
The urn model problem is a classic example used to illustrate probability concepts. In this problem, the variability of the system affects the outcome of events. We're not just drawing balls randomly, but altering the conditions of the next draw based on the first.
Adding the same-colored balls back into the urn changes the probability landscape: this models changing probabilities over sequential events.
  • In the first draw, the probabilities depend solely on the original counts \((b\) and \(r)\).
  • After adding balls, the second draw probabilities alter, changing the sample space.
Acknowledging these changes is crucial to understanding how they affect calculations further along.
Event probability calculation
Calculating event probability involves understanding and calculating these sequences of events. In this problem, events are connected:
  • Event A: drawing a black ball first.
  • Event C: drawing a red ball second.
Given the changes in the numbers of balls, we calculate the probability of the sequence (A then C) and the sequence (B then C). These probabilities famously derive from the conditions around the drawing, accounting for the changes.
To find \( P(A|C) \), we apply these concepts via Bayes' theorem, working through the conditional probabilities for each possible sequence, ultimately leading to the simplified solution. The thorough application of these calculations shows the depth of interconnectedness in probability theory applications.

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Most popular questions from this chapter

Let \(A\) and \(B\) be events having positive probability. State whether each of the following statements is (i) necessarily true, (ii) necessarily false, or (iii) possibly true. (a) If \(A\) and \(B\) are mutually exclusive, then they are independent. (b) If \(A\) and \(B\) are independent, then they are mutually exclusive. (c) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are mutually exclusive. (d) \(P(A)=P(B)=.6\), and \(A\) and \(B\) are independent.

Suppose that an ordinary deck of 52 cards (which contains 4 aces) is randomly divided into 4 hands of 13 cards each. We are interested in determining \(p\), the probability that each hand has an ace. Let \(E_{i}\) be the event that the \(i\) th hand has exactly one ace. Determine \(p=P\left(E_{1} E_{2} E_{3} E_{4}\right)\) by using the multiplication rule.

Urn I contains 2 white and 4 red balls, whereas urn II contains 1 white and 1 red ball. A ball is randomly chosen from urn I and put into um \(\Pi\), and a ball is then randomly selected from urn II. What is (a) the probability that the ball selected from urn II is white; (b) the conditional probability that the transferred ball was white, given that a white ball is selected from urn II?

In a certain species of rats, black dominates over brown. Suppose that a black rat with two black parents has a brown sibling. (a) What is the probability that this rat is a pure black rat (as opposed to being a hybrid with one black and one brown gene)? (b) Suppose that when the black rat is mated with a brown rat, all 5 of their offspring are black. Now, what is the probability that the rat is a pure black rat?

A true-false question is to be posed to a husband and wife team on a quiz show. Both the husband and the wife will, independently, give the correct answer with probability \(p\). Which of the following is a better strategy for this couple? (a) Choose one of them and let that person answer the question; or (b) have them both consider the question and then either give the common answer if they agree or, if they disagree, flip a coin to determine which answer to give?
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