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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

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

Again consider a Little League team that has 15 players on its roster. a. How many ways are there to select 9 players for the starting lineup? b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters? c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

The three most popular options on a certain type of new car are a built-in GPS \((A)\), a sunroof \((B)\), and an automatic transmission \((C)\). If \(40 \%\) of all purchasers request \(A, 55 \%\) request \(B, 70 \%\) request \(C, 63 \%\) request \(A\) or \(B, 77 \%\) request \(A\) or \(C, 80 \%\) request \(B\) or \(C\), and \(85 \%\) request \(A\) or \(B\) or \(C\), determine the probabilities of the following events. [Hint: " \(A\) or \(B\) " is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and not either of the other two options. d. The next purchaser will select exactly one of these three options.

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B\) ), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\) and \(P(A \cup B)\) ?

Individual A has a circle of five close friends (B, C, D, E, and F). A has heard a certain rumor from outside the circle and has invited the five friends to a party to circulate the rumor. To begin, A selects one of the five at random and tells the rumor to the chosen individual. That individual then selects at random one of the four remaining individuals and repeats the rumor. Continuing, a new individual is selected from those not already having heard the rumor by the individual who has just heard it, until everyone has been told. a. What is the probability that the rumor is repeated in the order \(\mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}\), and \(\mathrm{F}\) ? b. What is the probability that \(\mathrm{F}\) is the third person at the party to be told the rumor? c. What is the probability that \(\mathrm{F}\) is the last person to hear the rumor? d. If at each stage the person who currently "has" the rumor does not know who has already heard it and selects the next recipient at random from all five possible individuals, what is the probability that \(\mathrm{F}\) has still not heard the rumor after it has been told ten times at the party?

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