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91Ó°ÊÓ

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 in both of the next 2 years? All of the next 3 years?

Short Answer

Expert verified
The probability of an eligible person being selected for jury duty in both of the next 2 years is 0.0225 or 2.25%, and the probability of being selected in all of the next 3 years is 0.003375 or 0.3375%.

Step by step solution

01

Identify the probability of being selected for jury duty in one year

We are given that the probability of being selected for jury duty in one year is 15%. We can write this as a decimal probability: \(P(A) = 0.15\), where A represents being selected for jury duty in one year.
02

Identify the events as independent

Since the same individual cannot be called more than once in the same year, and people are selected at random from those eligible, we can assume that the events of being selected for jury duty in consecutive years are independent events.
03

Calculate the probability of being selected for jury duty in the next 2 years

To find the probability of being selected for jury duty in both of the next 2 years, we need to multiply the probability of being selected in one year by the probability of being selected in the following year since they are independent events: \(P(A_1 \cap A_2) = P(A_1) \cdot P(A_2)\) Here, \(A_1\) and \(A_2\) represent being selected for jury duty in the first and second year respectively. Since the probability of being selected in one year is 0.15, then: \(P(A_1 \cap A_2) = 0.15 \cdot 0.15\) \(P(A_1 \cap A_2) = 0.0225\) The probability of being selected for jury duty in both of the next 2 years is 0.0225 or 2.25%.
04

Calculate the probability of being selected for jury duty in the next 3 years

To find the probability of being selected for jury duty in all of the next 3 years, we need to multiply the probability of being selected in one year by the probability of being selected in the following two years, since they are independent events: \(P(A_1 \cap A_2 \cap A_3) = P(A_1) \cdot P(A_2) \cdot P(A_3)\) Here, \(A_1\), \(A_2\), and \(A_3\) represent being selected for jury duty in the first, second, and third year respectively. Since the probability of being selected in one year is 0.15, then: \(P(A_1 \cap A_2 \cap A_3) = 0.15 \cdot 0.15 \cdot 0.15\) \(P(A_1 \cap A_2 \cap A_3) = 0.003375\) The probability of being selected for jury duty in all of the next 3 years is 0.003375 or 0.3375%.

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Key Concepts

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

Understanding Independent Events in Probability
In probability theory, understanding the concept of independent events is fundamental. Independent events are two or more events where the occurrence of one event does not affect the occurrence of another. In simpler terms, the outcome of one event has no impact on the outcome of another. For instance, flipping a coin and rolling a die are independent events because the result of the coin flip does not influence the result of the die roll.

Applying this to the context of our exercise where individuals are called for jury duty, we can say that being selected for jury duty in one year is an independent event from being selected in another year. This is because the selection in one year does not affect the selection in the following year, given that the same individual cannot be selected more than once in a calendar year.

It's worth noting that identifying events as independent is critical before we can use certain rules to calculate the combined probability of these events. If events were not independent, we would need to approach the problem differently, perhaps with rules concerning conditional probabilities or dependent events.
Probability Calculation: The Basics
The probability calculation involves finding the likelihood of an event occurring. This is often expressed as a fraction or decimal between 0 and 1, where 0 indicates that the event will never occur, and 1 indicates certainty that the event will occur. For example, if the probability of an event is 0.5, this means there is a 'fifty-fifty' chance of the event occurring.

In our exercise, the probability that an eligible person is selected for jury duty in one year is given as 15%, which translates to 0.15 when expressed as a decimal. To comprehend probability calculations, one must convert percentages to decimals as such and understand that a 15% chance of occurring is the same as saying the event will occur 15 times out of 100 trials on average.

When we speak about calculating the probability of multiple independent events occurring sequentially, such as being selected for jury duty multiple years in a row, we move beyond simple probability calculation and into the realm of combined probabilities, which is governed by the multiplication rule.
Multiplication Rule for Independent Events
The multiplication rule for independent events is a vital tool in probability. It allows us to calculate the probability of two or more independent events occurring together by multiplying their separate probabilities. The formula for this rule is:\[ P(A \cap B) = P(A) \cdot P(B) \]where \( P(A \cap B) \) is the probability of both events \( A \) and \( B \) occurring, and \( P(A) \) and \( P(B) \) are the probabilities of each event occurring separately. This multiplication rule only applies when the events are independent, as in our jury duty scenario.

When we apply this rule to our example, we discover the combined probability of being selected for jury duty for consecutive years. For two years, the calculation would be \( 0.15 \cdot 0.15 \), and for three years, it would be \( 0.15 \cdot 0.15 \cdot 0.15 \). Thus, the multiplication rule simplifies the process of finding probabilities for a series of independent events, making it an indispensable component in a wide array of probability problems.

Consequently, grasping the multiplication rule is essential for anyone dealing with probability, because it frequently serves as the foundation for more complex probability theories and applications.

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Most popular questions from this chapter

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.64\) and \(P(F)=0.17\) i. What is the value of \(P(E \cap F)\) ? ii. What is the value of \(P(E \cup F)\) ? b. Suppose that \(A\) and \(B\) are events with \(P(A)=0.3, P(B)=0.5\), and \(P(A \cap B)=0.15 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell? c. Suppose that \(A\) and \(B\) are events with \(P(A)=0.65\) and \(P(B)=0.57 .\) Are \(A\) and \(B\) mutually exclusive? How can you tell?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=0.4\) and \(P\left(E_{2}\right)=0.3\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\operatorname{not} E_{1}\) and not \(E_{2}\) ). c. What is the probability that the firm is successful in at least one of the two bids?

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and one is assigned to Treatment 1 and the other to Treatment 2 . 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 by \(2 .\) For example, based on the results in the accompanying table, 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 0.7 and Treatment 2 has a success rate of \(0.4 .\) 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 represent 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 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. Estimate 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. Estimate the probability that the researchers will incorrectly conclude that Treatment 2 is the better treatment.

5.65 The authors of the paper "Do Physicians Know When Their Diagnoses Are Correct?" (Journal of General Internal Medicine [2005]: 334-339) presented detailed case studies to medical students and to faculty at medical schools. Each participant was asked to provide a diagnosis in the case and also to indicate whether his or her confidence in the correctness of the diagnosis was high or low. Define the events \(C\), \(I,\) and \(H\) as follows: \(C=\) event that diagnosis is correct \(I=\) event that diagnosis is incorrect \(H=\) event that confidence in the correctness of the diagnosis is high a. Data appearing in the paper were used to estimate the following probabilities for medical students: $$ \begin{array}{r} P(C)=0.261 \\ P(I)=0.739 \\ P(H \mid C)=0.375 \\ P(H \mid I)=0.073 \end{array} $$ Use Bayes' Rule to calculate the probability of a correct diagnosis given that the student's confidence level in the correctness of the diagnosis is high. b. Data from the paper were also used to estimate the following probabilities for medical school faculty: $$ \begin{array}{r} P(C)=0.495 \\ P(I)=0.505 \\ P(H \mid C)=0.537 \\ P(H \mid I)=0.252 \end{array} $$ Calculate \(P(C \mid H)\) for medical school faculty. How does the value of this probability compare to the value of \(P(C \mid H)\) for students calculated in Part (a)?

The probability of getting a king when a card is selected at random from a standard deck of 52 playing cards is \(\frac{1}{13}\). a. Give a relative frequency interpretation of this probability. b. Express the probability as a decimal rounded to three decimal places. Then complete the following statement: If a card is selected at random, I would expect to see a king about times in 1000 .

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