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Repeat Example \(1 \mathrm{c}\) when the balls are selected with replacement.

Short Answer

Expert verified
The probability of the desired sequence, say A-B-C, when the balls are selected with replacement is calculated as \(P(A-B-C) = P(A) \times P(B) \times P(C)\), where \(P(A) = \frac{n_A}{n_{total}}\), \(P(B) = \frac{n_B}{n_{total}}\), and \(P(C) = \frac{n_C}{n_{total}}\). Substitute the probabilities obtained and simplify the result as a fraction or decimal.

Step by step solution

01

Understand the problem

In this exercise, we have a box containing balls of different colors. The task is to calculate the probability of selecting balls of specific colors with replacement. With replacement means that after selecting a ball, it is placed back into the box before selecting the next ball.
02

Identify the colors and their frequencies

Determine which colors the balls are and how many of each color there is in the box. For this step, assume there are balls of color A, color B, and color C, and let \(n_A\), \(n_B\), and \(n_C\) represent their respective frequencies. The total number of balls is given by \(n_{total} = n_A + n_B + n_C\).
03

Calculate probability of each color

To calculate the probability of selecting each color of the ball, we need to use the formula for probability, which is: \(P(X) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\) For color A, the probability, \(P(A)\), would be: \(P(A) = \frac{n_A}{n_{total}}\) Similarly, for colors B and C, calculate the probabilities \(P(B)\) and \(P(C)\): \(P(B) = \frac{n_B}{n_{total}}\) \(P(C) = \frac{n_C}{n_{total}}\)
04

Identify the desired sequence of colors

In this step, specify the sequence of colors to be selected. Depending on the exercise, it could be any sequence, such as A-B-C, A-A-C, B-C-A, etc.
05

Calculate probability of the desired sequence

Now, calculate the probability of the specified sequence by multiplying the probabilities of each color being selected in that order. Since the balls are replaced after each selection, the probabilities for each color remain the same. For example, if the sequence is A-B-C, the probability of this sequence \(P(A-B-C)\) would be: \(P(A-B-C) = P(A) \times P(B) \times P(C)\)
06

Simplify and present the result

After calculating the probability, simplify the result by substituting the probabilities obtained in Step 3 and present the result as a simplified fraction or decimal as required.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with the likelihood of events occurring. It is the foundation for many disciplines including statistics, economics, and engineering. The essence of probability theory lies in quantifying uncertainty and making calculated predictions based on available data.

In the context of our exercise, probability theory helps us determine the chance of picking a ball of a certain color from a box when the balls are selected with replacement. To calculate this, you would use the formula:
\[\begin{equation} P(X) = \frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}}\end{equation}\]

Each time a ball is selected, it is considered an independent event because being picked once does not affect its chances of being picked again due to the replacement. This is a key concept in probability theory called independence, where the outcome of one event does not influence the outcome of another.
Combinatorics
Combinatorics is a field of mathematics primarily concerned with counting, both as a means and an end in obtaining results, and certain properties of finite structures. It includes a variety of topics, such as graph theory, enumeration, and design theory. One of its core principles is determining the number of ways specific outcomes can be achieved, which is fundamental in calculating probabilities.

Within our particular scenario, combinatorics comes into play when we identify the different sequences in which balls can be drawn from the box. For example, if the sequence we're considering is drawing a red ball followed by a green ball and then a blue ball, combinatorics helps to understand this specific arrangement. This is where the concept of permutations and combinations would typically come into play. However, when the balls are selected with replacement, the combinatorial calculations are simplified as each draw is independent and the total number of possibilities remains constant.
Conditional Probability
Conditional probability is a measure of the likelihood of an event occurring given that another event has already occurred. This concept is integral to understanding more complex probability scenarios where events are interdependent. The probability of event A given event B is notated as P(A|B).

In the exercise scenario, although we deal with replacement—meaning each event is independent from each other and hence does not fit the exact definition of conditional probability—it’s crucial to understand this concept to grasp more complicated probability problems where replacement is not involved. For instance, if the balls were not replaced, the probability of drawing a ball of color B after having already drawn a ball of color A would depend on the prior outcome and thus be a conditional probability. The computation involves adjusting the probability of B based on the occurrence of A. However, since our exercise is with replacement, each draw is independent, and the conditional probability in standard form is not required.

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Most popular questions from this chapter

Each of 500 soldiers in an army company independently has a certain disease with probability \(1 / 10^{3} .\) This disease will show up in a blood test, and to facilitate matters, blood samples from all 500 soldiers are pooled and tested. (a) What is the (approximate) probability that the blood test will be positive (that is, at least one person has the disease)? Suppose now that the blood test yields a positive result. (b) What is the probability, under this circumstance, that more than one person has the disease? Now, suppose one of the 500 people is Jones, who knows that he has the disease. (c) What does Jones think is the probability that more than one person has the disease? Because the pooled test was positive, the authorities have decided to test each individual separately. The first \(i-1\) of these tests were negative, and the \(i\) th one-which was on Jones-was positive. (d) Given the preceding scenario, what is the probability, as a function of \(i,\) that any of the remaining people have the disease?

In some military courts, 9 judges are appointed. However, both the prosecution and the defense attorneys are entitled to a peremptory challenge of any judge, in which case that judge is removed from the case and is not replaced. A defendant is declared guilty if the majority of judges cast votes of guilty, and he or she is declared innocent otherwise. Suppose that when the defendant is, in fact, guilty, each judge will (independently) vote guilty with probability. \(7,\) whereas when the defendant is, in fact, innocent, this probability drops to .3. (a) What is the probability that a guilty defendant is declared guilty when there are (i) \(9,\) (ii) \(8,\) and (iii) 7 judges? (b) Repeat part (a) for an innocent defendant. (c) If the prosecuting attorney does not exercise the right to a peremptory challenge of a judge, and if the defense is limited to at most two such challenges, how many challenges should the defense attorney make if he or she is 60 percent certain that the client is guilty?

Two fair dice are rolled. Let \(X\) equal the product of the 2 dice. Compute \(P\\{X=i\\}\) for \(i=1, \ldots, 36\)

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all \(10 !\) possible rankings are equally likely. Let \(X\) denote the highest ranking achieved by a woman. (For instance, \(X=1\) if the top-ranked person is female.) Find \(\mathrm{P}\\{\mathrm{X}=\mathrm{i}\\}, i=1,2,3, \ldots, 8,9,10\).

A communications channel transmits the digits 0 and 1. However, due to static, the digit transmitted is incorrectly received with probability .2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of \(1 .\) If the receiver of the message uses "majority" decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making?

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