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In a small city, approximately \(15 \%\) of those eligible are called for jury duty in any one calendar year. People are selected for jury duty at random from those eligible, and the same individual cannot be called more than once in the same year. What is the probability that a particular eligible person in this city is selected in each of the next 2 years? In each of the next 3 years?

Short Answer

Expert verified
The probability that a particular eligible person in this city is selected in each of the next 2 years is \(0.0225\) (or \(2.25%\) ) and in each of the next 3 years is \(0.003375\) (or \(0.3375%\) ).

Step by step solution

01

Define the probability of being selected for jury duty

It is given that 15% of eligible persons are chosen for jury duty in a year. The probability of getting chosen is given by this percentage, which is equal to 0.15
02

Calculate the probability for two consecutive years

The person is chosen independently in each year. Therefore, the probability of being chosen in each of two consecutive years is simply the product of the probabilities for each year. So, it is \(0.15 * 0.15 = 0.0225\)
03

Same calculation for three consecutive years

By the same logic, the probability of being chosen in each of three consecutive years is the product of the probabilities for each of these years. So, it is \(0.15 * 0.15 * 0.15 = 0.003375\)

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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 calculating the likelihood of various outcomes. It helps us quantify uncertainty and make informed predictions about events that have random outcomes. In probability theory, the outcomes could be anything from the roll of a dice to the selection for jury duty in a city, as seen in the problem we are discussing.

For instance, when we say there is a 15% chance of an eligible person being called for jury duty in the scenario provided, we are using probability theory to state that if we were to randomly select many groups of eligible people, approximately 15 out of every 100 would be expected to be called for jury duty in any one year. Probability is calculated as a number between 0 and 1, where 0 indicates an impossibility, and 1 denotes a certainty.
Consecutive Events Probability
When dealing with consecutive events probability, we are interested in the chances of multiple outcomes happening in a sequence. If the events are independent, which means the outcome of one event does not impact the other, then the total probability is the product of the individual probabilities.

Our example problem illustrates this beautifully. Since being selected for jury duty in one year does not affect the selection in the following years, the probabilities can be multiplied together. Hence, for two consecutive years, we multiply the annual probability of being selected (\(0.15\) for each year) to find the probability for both years. This multiplication rule simplifies analyses involving independent consecutive events greatly.
Random Selection
Random selection is a method by which subjects are chosen randomly from a larger pool such that each subject has an equal chance of being selected. It is a cornerstone concept for ensuring fairness and objectivity in various procedures, including lotteries and, as in our example, jury duty selection.

In the context of our problem, every eligible person has an equal chance of 15% to be selected for jury duty each year, ensuring that the same individual is not called upon again in the same year. This process ensures that over a long period, the jury duty selection remains fair and not biased toward any subset of individuals.
Statistical Analysis
Statistical analysis involves collecting, examining, summarizing, and interpreting data to uncover patterns and trends. This type of analysis is useful in understanding the behaviors of complex systems and making decisions based on data rather than guesses.

In the exercise we are looking at, statistical analysis would be used to monitor the selection process of jurors to ensure that it remains at the stated probability of 15%. Over the long term, statistical tools could help identify if any anomalies or biases exist in the system. This way, those responsible for the jury selection process can make adjustments to maintain the integrity and randomness of the selection process.

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Most popular questions from this chapter

Five hundred first-year students at a state university were classified according to both high school GPA and whether they were on academic probation at the end of their first semester. The data are $$ \begin{array}{lcccc} && {\text { High School GPA }} \\ & 2.5 \text { to } & 3.0 \text { to } & 3.5 \text { and } & \\ \text { Probation } & <3.0 & <3.5 & \text { Above } & \text { Total } \\ \hline \text { Yes } & 50 & 55 & 30 & 135 \\ \text { No } & 45 & 135 & 185 & 365 \\ \text { Total } & 95 & 190 & 215 & 500 \\ \hline \end{array} $$ a. Construct a table of the estimated probabilities for each GPA-probation combination. b. Use the table constructed in Part (a) to approximate the probability that a randomly selected first-year student at this university will be on academic probation at the end of the first semester. c. What is the estimated probability that a randomly selected first-year student at this university had a high school GPA of \(3.5\) or above? d. Are the two outcomes selected student has a bigh school GPA of \(3.5\) or above and selected student is on academic probation at the end of the first semester independent outcomes? How can you tell? e. Estimate the proportion of first-year students with high school GPAs between \(2.5\) and \(3.0\) who are on academic probation at the end of the first semester. f. Estimate the proportion of those first-year students with high school GPAs \(3.5\) and above who are on academic probation at the end of the first semester.

Delayed diagnosis of cancer is a problem because it can delay the start of treatment. The paper "Causes of Physician Delay in the Diagnosis of Breast Cancer" (Archives of Internal Medicine \([2002]: 1343-1348)\) examined possible causes for delayed diagnosis for women with breast cancer. The accompanying table summarizes data on the initial written mammogram report (benign or suspicious) and whether or not diagnosis was delayed for 433 women with breast cancer. $$ \begin{array}{l|cc} & & \text { Diagnosis } \\ & \begin{array}{c} \text { Diagnosis } \\ \text { Delayed } \end{array} & \begin{array}{c} \text { Not } \\ \text { Delayed } \end{array} \\ \hline \begin{array}{l} \text { Mammogram Report Benign } \\ \text { Mammogram Report } \\ \text { Suspicious } \end{array} & 32 & 89 \\ & 8 & 304 \\ \hline \end{array} $$ Consider the following events: \(B=\) the event that the mammogram report says benign \(S=\) event that the mammogram report says suspicious \(D=\) event that diagnosis is delayed a. Assume that these data are representative of the larger group of all women with breast cancer. Use the data in the table to find and interpret the following probabilities: i. \(\quad P(B)\) ii. \(P(S)\) iii. \(P(D \mid B)\) iv. \(P(D \mid S)\) b. Remember that all of the 433 women in this study actually had breast cancer, so benign mammogram reports were, by definition, in error. Write a few sentences explaining whether this type of error in the reading of mammograms is related to delayed diagnosis of breast cancer.

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over 6 feet in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A)\) ? Why?

The following case study was reported in the article “Parking Tidkets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the Unknown. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=\) \(1 / 144\). a. What reasoning did the expert use to arrive at the probability of \(1 / 144\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 144 ?\) What effect does this error have on the probability of occurrence? Do you think that \(1 / 144\) is larger or smaller than the correct probability of occurrence?

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)

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