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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
Probability of red-red: \(\frac{38}{55}\); Original configuration: \(\frac{30}{55}\).

Step by step solution

01

Calculate Probability of Red Ball from First Box

The first box contains 6 red balls and 4 green balls. The probability of picking a red ball from this box is given by the ratio of red balls to the total balls:\[P(R_1) = \frac{6}{10} = 0.6\]
02

Calculate Probability of Red Ball from Second Box (Case 1)

If a red ball is transferred to the second box, it then contains 8 red balls and 3 green balls. The probability of selecting a red ball from the second box in this scenario is:\[P(R_2|R_1) = \frac{8}{11}\]
03

Calculate Probability of Red Ball from Second Box (Case 2)

If a green ball is transferred to the second box, it contains 7 red balls and 4 green balls. The probability of selecting a red ball from the second box in this case is:\[P(R_2|G_1) = \frac{7}{11}\]
04

Combine Probabilities for Both Scenarios

The total probability of selecting a red ball first from the first box and then a red ball from the second box can be calculated using the law of total probability:\[P(R_1 \cap R_2) = P(R_1) \cdot P(R_2|R_1) + P(G_1) \cdot P(R_2|G_1)\]Where \(P(G_1) = \frac{4}{10} = 0.4\). Substituting:\[P(R_1 \cap R_2) = 0.6 \times \frac{8}{11} + 0.4 \times \frac{7}{11} = \frac{48}{110} + \frac{28}{110} = \frac{76}{110} = \frac{38}{55}\]
05

Calculate Probability of Original Ball Numbers

For the numbers of red and green balls in the first box to remain unchanged,- A red ball is selected first, then a red ball from the second box.- A green ball is selected first, then a green ball from the second box.The overall probability is:\[P(\text{original}) = (0.6 \times \frac{8}{11}) + (0.4 \times \frac{3}{11}) = \frac{48}{110} + \frac{12}{110} = \frac{60}{110} = \frac{30}{55}\]

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Key Concepts

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

Law of Total Probability
Understanding the Law of Total Probability is crucial for solving problems involving multiple possible outcomes. This law helps us determine the probability of an event by considering all possible ways the event can happen. Essentially, it breaks down complex problems into simpler parts by examining different scenarios or cases.

In the given exercise, we deal with two different scenarios:
  • A red ball is moved from the first box to the second box.
  • A green ball is moved from the first box to the second box.

By considering these scenarios separately, we can calculate the overall probability of picking a red ball from both boxes. This is done by multiplying the probability of each scenario by its respective probability of occurrence and then adding these products together. This method is neatly wrapped up in the equation:
\[P(R_1 \cap R_2) = P(R_1) \times P(R_2|R_1) + P(G_1) \times P(R_2|G_1)\]
The law of total probability therefore serves as a fundamental tool for breaking down complex probability problems involving multiple steps or actions.
Probability Theory
Probability Theory is the mathematical framework that allows us to make predictions about future events based on a known model of a random process. In simple terms, it's the tool we use to understand how likely different outcomes are.

For the exercise, we worked with the basic definitions of probability. To find out the probability of selecting different colored balls, we looked at the ratio of the desired outcomes (like picking a red ball) over the total possible outcomes. For example:
  • The probability of selecting a red ball from the first box is calculated as \( \frac{6}{10} \).
  • This is the ratio of red balls to the total balls in that box.

Probability can get complex when events are dependent, like when the outcome of one event affects another. This is important here as the movement of balls between boxes changes the probabilities we need to calculate for subsequent picking events. Therefore, probability theory helps us make logical predictions about the processes by examining these complex dependencies and scenarios.
Combinatorial Analysis
Combinatorial Analysis deals with counting, arranging, and analyzing configurations of sets, especially when it comes to questions about how transitions or exchanges affect overall probability. It's a versatile tool often used alongside probability theory to solve counting problems.

In this exercise, combinatorial analysis aids in understanding the different configurations of balls in each box after one ball is moved to another. Our approach is grounded in recognizing that picking and exchanging events can vastly change these configurations. For instance, knowing how a single red or green ball can change the contents of the second box leads to a clearer calculation of subsequent probabilities.

This analysis doesn't just count possibilities, it considers how each move impacts further actions. It is crucial when determining how possible outcomes develop through each step of the process, informing the decisions in calculating total probabilities. Thus, combinatorial analysis complements probability theory by addressing the counting and configuration aspects in these complex probability problems.

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Most popular questions from this chapter

a. A lumber company has just taken delivery on a lot of \(10,0002 \times 4\) boards. Suppose that \(20 \%\) of these boards \((2,000)\) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let \(A=\\{\) the first board is green \(\\}\) and \(B=\\{\) the second board is green \(\\}\). Compute \(P(A), P(B)\), and \(P(A \cap B)\) (a tree diagram might help). Are \(A\) and \(B\) independent? b. With \(A\) and \(B\) independent and \(P(A)=P(B)=.2\), what is \(P(A \cap B)\) ? How much difference is there between this answer and \(P(A \cap B)\) in part (a)? For purposes of calculating \(P(A \cap B)\), can we assume that \(A\) and \(B\) of part (a) are independent to obtain essentially the correct probability? c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for \(P(A \cap B)\) ? What is the critical difference between the situation here and that of part (a)? When do you think an independence assumption would be valid in obtaining an approximately correct answer to \(P(A \cap B)\) ?

Each of a sample of four home mortgages is classified as fixed rate \((F)\) or variable rate \((V)\). a. What are the 16 outcomes in \(\mathcal{S}\) ? b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate? c. Which outcomes are in the event that all four mortgages are of the same type? d. Which outcomes are in the event that at most one of the four is a variable- rate mortgage? e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events? f. What are the union and intersection of the two events in parts (b) and (c)?

An individual is presented with three different glasses of cola, labeled \(C, D\), and \(P\). He is asked to taste all three and then list them in order of preference. Suppose the same cola has actually been put into all three glasses. a. What are the simple events in this ranking experiment, and what probability would you assign to each one? b. What is the probability that \(C\) is ranked first? c. What is the probability that \(C\) is ranked first and \(D\) is ranked last?

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent.

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries. a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this? b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this? c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety? d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen? e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

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