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Jeanie is a bit forgetful, and if she doesn't make a "to do" list, the probability that she forgets something she is supposed to do is 1 . Tomorrow she intends to run three errands, and she fails to write them on her list. a. What is the probability that Jeanie forgets all three errands? What assumptions did you make to calculate this probability? b. What is the probability that Jeanie remembers at least one of the three errands? c. What is the probability that Jeanie remembers the first errand but not the second or third?3

Short Answer

Expert verified
a. The probability that Jeanie forgets all three errands is 1. b. The probability that Jeanie remembers at least one of the three errands is 0. c. The probability that Jeanie remembers the first errand but not the second or third is 1.

Step by step solution

01

Assuming Independence

Firstly, let's proceed by assuming each errand is an independent event. In other words, forgetting or remembering each errand doesn't affect the probabilities of forgetting/remembering the other errands. This assumption is implicit as it's not mentioned the opposite in the exercise.
02

Calculate the probability of forgetting all 3 errands

Since we're finding the probability of Jeanie forgetting every errand, and assuming each errand is an independent event, the calculation is straightforward. The probability of forgetting something is given as 1, hence the probability that she forgets all three errands is \(1 \times 1 \times 1 = 1\).
03

Calculate the probability of remembering at least 1 errand

The question asks us to calculate the probability that at least one errand is remembered. This is actually the complementary event to the event that all errands are forgotten. So the probability that at least one errand is remembered is \(1 - \) Probability of forgetting all three errands = \(1 - 1 = 0\).
04

Calculate the probability of remembering only the first errand

The question asks us to find the probability that she remembers the first errand but forgets the others. Since each errand is an independent event, we multiply the probabilities together. Still, the probability is \(1 \times 1 \times 1 = 1\).

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Key Concepts

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

Independent Events
When considering probability, an independent event is one where the occurrence of one event does not affect the probability of another event occurring. In other words, independent events have no influence on each other. For instance, rolling a die and flipping a coin are independent events because the result of the die roll does not change the chances of getting heads or tails on the coin flip.

This concept is crucial in the exercise involving Jeanie and her errands. The problem assumes that whether Jeanie remembers or forgets to do each errand is independent of the others. Hence, the probability of her forgetting any single task does not impact the probability of forgetting the next. When calculating the probability of multiple independent events occurring, we multiply the probabilities of each event. However, there's a point of improvement to be made here since the solution assumes a forgetting probability of 1 (or 100%) for each errand, which is unrealistic in typical scenarios and should be adjusted to reflect a likelihood less than 1.
Complementary Events
The concept of complementary events is applied when we want to know the probability of an event not happening. For every event A, there is a complementary event that encompasses all other outcomes than A. The probability of an event and its complement always sum up to 1. This is a simple yet powerful concept as it allows us to calculate probabilities indirectly.

For instance, if the probability of it raining tomorrow is 0.3, then the probability that it does not rain is 0.7, because these two probabilities must add up to 1. In the context of our exercise, the solution calculates the probability of Jeanie remembering at least one errand by using the complement of the probability of Jeanie forgetting all errands. However, this approach has assumed an incorrect calculation in Step 2, where the forgetting probability should not be 1 for each event, thus the complement calculation in Step 3 is also incorrect and needs correction.
Probability Calculations
Performing accurate probability calculations is essential for solving problems in probability theory. To calculate the probability of a single event, you divide the number of favorable outcomes by the total number of possible outcomes. For multiple events, especially independent ones, you would multiply the probabilities of the individual events, as per the multiplication rule.

Returning to our example with Jeanie, we would ideally determine the probability of her forgetting an errand, which should be less than 1, and then use this figure to calculate the compound probability of independent events. Furthermore, if we had the correct probabilities, we would use the addition rule to determine the probability of remembering at least one errand, which requires calculating the probability of the event's complement, as addressed in Step 3 of the solution. By ensuring the use of accurate probabilities and proper calculation methods, we can derive meaningful conclusions from our probability problems.

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Most popular questions from this chapter

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1, the players seeded (rated) first and fourth play. In Game 2, the players seeded second and third play. In Game 3 , the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are given: $$\begin{aligned} &P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=.8 \\ &P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 2)=.6 \\ &P(\text { Seed } 1 \text { defeats Seed } 3)=.7 \\ &P(\text { Seed } 2 \text { defeats } \text { Seed } 3)=.6 \\ &P(\text { Seed } 2 \text { defeats Seed } 4)=.7 \\ &P(\text { Seed } 3 \text { defeats Seed } 4)=.6 \end{aligned}$$ a. Describe how you would use a selection of random digits to simulate Game 1 of this tournament. b. Describe how you would use a selection of random digits to simulate Game 2 of this tournament. c. How would you use a selection of random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the true probability? Explain.

Students at a particular university use a telephone registration system to select their courses for the next term. There are four different priority groups, with students in Group 1 registering first, followed by those in Group 2, and so on. Suppose that the university provided the accompanying information on registration for the fall semester. The entries in the table represent the proportion of students falling into each of the 20 priority-unit combinations. $$\begin{array}{c|ccccc} \begin{array}{c} \text { Priority } \\ \text { Group } \end{array} & \mathbf{0 - 3} & \mathbf{4 - 6} & \mathbf{7 - 9} & \mathbf{1 0 - 1 2} & \begin{array}{c} \text { More } \\ \text { Than 12 } \end{array} \\ \hline \mathbf{1} & .01 & .01 & .06 & .10 & .07 \\ \mathbf{2} & .02 & .03 & .06 & .09 & .05 \\ \mathbf{3} & .04 & .06 & .06 & .06 & .03 \\ \mathbf{4} & .04 & .08 & .07 & .05 & .01 \\ & & & & & \end{array}$$ a. What proportion of students at this university got 10 or more units during the first call? b. Suppose that a student reports receiving 11 units during the first call. Is it more likely that he or she is in the first or the fourth priority group? c. If you are in the third priority group next term, is it likely that you will get more than 9 units during the first call? Explain.

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9\), \(P(2\) works \()=.9, P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that these four outcomes are independent (the four components work independently of one another). a. The \(1-2\) subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem was added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

\(6.23\) A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by \(2 .\) For example, they might observe the results in the table on the next page. The chance experiment would stop after the sixth pair, because Treatment 1 has two more successes than Treatment \(2 .\) Treatment 1 and success for Treatment 2 . Continue to select pairs, keeping track of the cumulative number of successes for each treatment. Stop the trial as soon as the number of successes for one treatment exceeds that for the other by 2 . This would complete one trial. Now repeat this whole process until you have results for at least 20 trials [more is better]. Finally, use the simulation results to estimate the desired probabilities.) a. What is the probability that more than 5 pairs must be treated before a conclusion can be reached? (Hint: \(P\) (more than 5\()=1-P(5\) or fewer \() .)\) b. What is the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment?

The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. Postal Service standards call for overnight delivery within a zone of about \(60 \mathrm{mi}\) for any first-class letter deposited by the last collection time posted on a mailbox. Two-day delivery is promised within a 600 -mi zone, and three-day delivery is promised for distances over \(600 \mathrm{mi}\). The Price Waterhouse accounting firm conducted an independent audit by "seeding" the mail with letters and recording on-time delivery rates for these letters. Suppose that the results of the Price Waterhouse study were as follows (these numbers are fictitious, but they are compatible with summary values given in the article): $$\begin{array}{lcc} & \begin{array}{c} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Letters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \end{array}$$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that both of two letters mailed in New York are delivered on time d. The probability of on-time delivery nationwide

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