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Chapter 2: Problem 51

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

Short Answer

Expert verified
a. \( \frac{38}{55} \) b. \( \frac{32}{55} \)

Step by step solution

01

Calculate the probability of drawing a red ball from the first box

The first box contains 6 red balls and 4 green balls, totaling 10 balls. The probability of selecting a red ball from the first box is given by the ratio of red balls to the total number of balls: \( P(R_1) = \frac{6}{10} = \frac{3}{5} \).
02

Calculate the conditional probability of drawing a red ball from the second box, given a red ball was added

If a red ball is transferred from the first box to the second box, the second box will then contain 8 red balls and 3 green balls, totaling 11 balls. The probability of drawing a red ball from the second box after this transfer is \( P(R_2 | R_1) = \frac{8}{11} \).
03

Calculate the probability of drawing a green ball from the first box

The probability of selecting a green ball from the first box is \( P(G_1) = \frac{4}{10} = \frac{2}{5} \).
04

Calculate the conditional probability of drawing a red ball from the second box given a green ball is added

If a green ball is transferred to the second box, it will have 7 red balls and 4 green balls, totaling 11 balls. The probability of drawing a red ball now is \( P(R_2 | G_1) = \frac{7}{11} \).
05

Combine probabilities for question a

The probability that a red ball is selected from both boxes can be calculated using the law of total probability: \[ P(R_1 \text{ and } R_2) = P(R_1) \cdot P(R_2 | R_1) + P(G_1) \cdot P(R_2 | G_1) = \frac{3}{5} \cdot \frac{8}{11} + \frac{2}{5} \cdot \frac{7}{11} = \frac{24}{55} + \frac{14}{55} = \frac{38}{55}. \]
06

Determine the probability of the first box returning to its original state

The first box will return to its original state if either a red ball transfers both times or a green ball transfers both times. The probability for a red ball exchange is \( P(R_1 \text{ and } R_2) = \frac{3}{5} \cdot \frac{8}{11} = \frac{24}{55} \). The probability of a green ball sequence is \( P(G_1 \text{ and } G_2) = \frac{2}{5} \cdot \frac{4}{11} = \frac{8}{55} \). The total probability for the box state returning is \[ P(\text{original state}) = \frac{24}{55} + \frac{8}{55} = \frac{32}{55}. \]

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Key Concepts

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

Conditional Probability
Conditional probability helps us understand how the probability of an event changes when we have more information. It is denoted as \( P(A | B) \), which reads as the probability of event \( A \) occurring given that event \( B \) has already occurred. In our exercise, we calculate probabilities like \( P(R_2 | R_1) \), which means the probability of picking a red ball from the second box given that a red ball was initially transferred from the first box.
  • First, we determine how the initial event affects the overall scenario, such as moving a red or green ball from the first to the second box.
  • Then, we update our calculations for outcomes based on this new setup, considering the addition or removal of a ball.
This way of calculating gives us a better understanding of dependent events and how the initial conditions impact the chances of future outcomes.
Law of Total Probability
When dealing with multiple scenarios, the Law of Total Probability assists in finding the probability of an event by taking into account all possible ways it can happen. This law is very useful when our events are partitioned into several exclusive outcomes. In the problem, when trying to find the probability that a red ball is selected from both boxes, we utilize this law by considering all possible paths of transferring balls:
  • One path is transferring a red ball followed by picking another red from the second box, calculated as \( P(R_1) \times P(R_2 | R_1) \).
  • The second path is transferring a green, then picking a red, which is \( P(G_1) \times P(R_2 | G_1) \).
To get the final probability of selecting a red ball from both boxes, we sum the probabilities of all these paths, thus considering all possible combinations of transfers. It's a way to comprehensively account for all scenarios and their probabilities.
Combinatorics
Combinatorics is the field of mathematics focused on counting, arranging, and grouping objects. This is crucial in calculating probabilities, especially in determining the number of ways certain events can occur. Though not directly calculated in the exercise, understanding combinatorics helps in visualizing scenarios such as:
  • How many ways can a ball be chosen from a group?
  • What are the possible combinations of red and green balls in each pick?
By organizing and counting the possible outcomes systematically, the calculation of probability becomes more efficient. This is particularly useful when dealing with larger groups of objects or more complex probability scenarios, allowing us to enumerate possible events easily.
Random Variables
Random variables are a way to quantify random processes, assigning numerical values to the outcomes of a random phenomenon. In probability theory, a random variable represents how outcomes of different probabilities can be modeled. In our exercise context:
  • The selection of balls can be considered as random variables where each draw, say selecting a red or green ball, varies numerically.
  • Setting these outcomes as random variables helps us manage and calculate probabilities by providing a structured framework.
Random variables enable us to use mathematical tools and definitions, like probability distributions, to further analyze and predict the behavior of the selections being made. By utilizing random variables, we can more easily manipulate and interpret the random nature of these draws in probabilistic terms.

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Most popular questions from this chapter

A box contains the following four slips of paper, each having exactly the same dimensions: (1) win prize \(1 ;(2)\) win prize \(2 ;(3)\) win prize \(3 ;(4)\) win prizes 1,2 , and 3 . One slip will be randomly selected. Let \(A_{1}=\\{\) win prize 1\(\\}, A_{2}=\\{\) win prize 2\(\\}\), and \(A_{3}=\\{\) win prize 3\(\\} .\) Show that \(A_{1}\) and \(A_{2}\) are independent, that \(A_{1}\) and \(A_{3}\) are independent, and that \(A_{2}\) and \(A_{3}\) are also independent (this is pairwise independence). However, show that \(P\left(A_{1} \cap A_{2} \cap A_{3}\right) \neq P\left(A_{1}\right) \cdot P\left(A_{2}\right) \cdot P\left(A_{3}\right)\), so the three events are not mutually independent.

An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability. a. If \(20 \%\) of all seams need reworking, what is the probability that a rivet is defective? b. How small should the probability of a defective rivet be to ensure that only \(10 \%\) of all seams need reworking?

An academic department with five faculty members narrowed its choice for department head to either candidate \(A\) or candidate \(B\). Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for \(A\) and two for \(B\). If the slips are selected for tallying in random order, what is the probability that \(A\) remains ahead of \(B\) throughout the vote count (for example, this event occurs if the selected ordering is \(A A B A B\), but not for \(A B B A A)\) ?

At a gas station, \(40 \%\) of the customers use regular gas \(\left(A_{1}\right), 35 \%\) use mid-grade gas \(\left(A_{2}\right)\), and \(25 \%\) use premium gas \(\left(A_{3}\right)\). Of those customers using regular gas, only \(30 \%\) fill their tanks (event \(B\) ). Of those customers using mid- grade gas, \(60 \%\) fill their tanks, whereas of those using premium, \(50 \%\) fill their tanks. a. What is the probability that the next customer will request mid-grade gas and fill the tank \(\left(A_{2} \cap B\right)\) ? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? midgrade gas? Premium gas?

Seventy percent of the light aircraft that disappear while in flight in a certain country are subsequently discovered. Of the aircraft that are discovered, \(60 \%\) have an emergency locator, whereas \(90 \%\) of the aircraft not discovered do not have such a locator. Suppose a light aircraft has disappeared. a. If it has an emergency locator, what is the probability that it will not be discovered? b. If it does not have an emergency locator, what is the probability that it will be discovered?
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