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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 an eligible person in this city is selected 2 years in a row? 3 years in a row?

Short Answer

Expert verified
The probability that an eligible person in this city is selected 2 years in a row is approximately 2.25%, and for 3 years in a row is around 0.34%.

Step by step solution

01

Calculate the probability of being chosen one year

In this case, the problem states that the probability of an eligible person being chosen for jury duty any given year is 15%. Hence, the probability for one year, which can be symbolized as P(1 year), is 0.15.
02

Calculate the probability of being chosen two years in a row

To determine the probability that a person will be chosen two years in a row, we will multiply the probability of being chosen in the first year by the probability of being chosen in the second year (since these are independent events). Using the multiplication rule for independent events, where P(A and B) = P(A)P(B), we find P(2 years) = P(1 year) * P(1 year) = \(0.15 * 0.15\) which equals 0.0225 or 2.25%.
03

Calculate the probability of being chosen three years in a row

To determine the probability that a person will be chosen three years in a row, we multiply the probability of being chosen in the first year by the probability of being chosen in the second year and the probability of being chosen in the third year (since these are independent events). Using the multiplication rule for independent events, we find P(3 years) = P(1 year) * P(1 year) * P(1 year) = \(0.15 * 0.15 * 0.15\) = 0.003375 or around 0.34%

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Key Concepts

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

Independent Events
When we talk about independent events in probability, we are referring to situations where the occurrence of one event doesn't affect the probability of another. In simpler words, the events are not connected in any way.
For instance, in the context of the jury duty exercise, being chosen for jury duty in one year does not influence the chances of being selected in the following year.
This is because the selection is random and each selection is made afresh from all eligible candidates.
  • The key feature of independent events is that their probabilities multiple directly without one affecting the other.
  • This concept makes calculating probabilities over multiple trials straightforward using mathematical formulas.
Just keep in mind that all probabilities in such problems are calculated separately and then combined, reflecting their independence.
Multiplication Rule
The multiplication rule is a fundamental principle in probability used to find the probability of two or more independent events happening together.
It is expressed mathematically as:\[P(A \text{ and } B) = P(A) \cdot P(B)\]
  • "P(A and B)" signifies the probability of both events A and B occurring.
  • If the events are independent, this rule simplifies the joint probability calculation by simply multiplying their individual probabilities.
This rule was directly applied in the jury duty exercise to determine the likelihood of being selected multiple years in a row. It's essential to correctly identify the independence of events to use this rule accurately.
Jury Duty
Jury duty is a civic responsibility where eligible citizens are called to serve as jurors in courtroom trials. This process ensures that peers from the community are part of the legal system, helping to navigate and decide judicial outcomes. Selection is typically random, and eligibility varies by jurisdiction.
In our exercise, around 15% of eligible people are called each year. This means the pool of jurors is significant, and each individual has an equal chance of being called each year.
This random selection underpins the independent event framework discussed earlier.
Understanding this process helps contextualize the probability question, making the exercise relatable and relevant.
Probability Calculation
Probability calculation in the context of independent events, like jury duty selection, involves straightforward arithmetic using the multiplication rule. We calculate the chance an event happens by assessing each factor one at a time.
  • First, understand the base probability of a single event occurring, here being the juror selection in a single year (15% or 0.15).
  • Then, apply the multiplication rule over multiple years: for two years in a row, multiply the year's probability by itself: \(0.15 \times 0.15 = 0.0225\) or 2.25%.
  • For three consecutive years, multiply this result by the yearly probability again: \(0.15 \times 0.15 \times 0.15 = 0.003375\), roughly 0.34%.
This step-by-step approach simplifies complex probability questions by breaking them down into manageable parts. Understanding the probability principles ensures that calculations are logical and understandable.

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Most popular questions from this chapter

Many fire stations handle emergency calls for medical assistance as well as calls requesting firefighting equipment. A particular station says that the probability that an incoming call is for medical assistance is .85. This can be expressed as \(P(\) call is for medical assistance \()=.85 .\) a. Give a relative frequency interpretation of the given probability. b. What is the probability that a call is not for medical assistance? c. Assuming that successive calls are independent of one another (i.e., knowing that one call is for medical assistance doesn't influence our assessment of the probability that the next call will be for medical assistance), calculate the probability that both of two successive calls will be for medical assistance. d. Still assuming independence, calculate the probability that for two successive calls, the first is for medical assistance and the second is not for medical assistance. e. Still assuming independence, calculate the probability that exactly one of the next two calls will be for medical assistance. (Hint: There are two different possibilities that you should consider. The one call for medical assistance might be the first call, or it might be the second call.) f. Do you think it is reasonable to assume that the requests made in successive calls are independent? Explain.

Five hundred first-year students at a state university were classified according to both high school grade point average (GPA) and whether they were on academic probation at the end of their first semester. The data are summarized in the accompanying table. \begin{tabular}{lcccc} & \multicolumn{3}{c} { High School GPA } & \\ \cline { 2 - 4 } Probation & 2.5 to & 3.0 to & 3.5 and & \\ & \(<3.0\) & \(<3.5\) & Above & Total \\ \hline Yes & 50 & 55 & 30 & 135 \\ No & 45 & 135 & 185 & 365 \\ Total & 95 & 190 & 215 & 500 \\ \hline \end{tabular} a. Construct a table of the estimated probabilities for each GPA-probation combination by dividing the number of students in each of the six cells of the table by \(500 .\) 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 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.

A Gallup survey of 2002 adults found that \(46 \%\) of women and \(37 \%\) of men experience pain daily (San Luis Obispo Tribune, April 6,2000 ). Suppose that this information is representative of U.S. adults. If a U.S. adult is selected at random, are the outcomes selected adult is male and selected adult experiences pain daily independent or dependent? Explain.

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 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 at the bottom of the page. The experiment would stop after the sixth pair, because Treatment 1 has two more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment 2 . Suppose that Treatment 1 has a success rate of .7 (that is, \(P(\) success \()=.7\) for Treatment 1\()\) and that Treatment 2 has a success rate of .4. Use simulation to estimate the probabilities requested in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a success and \(8,9,\) and 0 to indicate a failure. Let the second digit represcnt Treatment \(2,\) with \(1-4\) representing a success. For example, if the two digits selected to represent a pair were 8 and 3 , you would record failure for Treatment 1 and success for Treatment 2. Continue to sclect pairs, kecping 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 five 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?

Three friends \((A, B,\) and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8\), and \(P(B\) beats \(C)=.6\) and that the outcomes of the three matches are independent of one another. a. What is the probability that \(A\) wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that A wins both her matches? c. What is the probability that A loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen. Calculate the probability of each separately, and then add.)

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