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Chapter 5: Problem 28

Each time a class meets, the professor selects one student at random to explain the solution to a homework problem. There are 40 students in the class, and no one ever misses class. Luke is one of these students. What is the probability that Luke is selected both of the next two times that the class meets?

Short Answer

Expert verified
The probability of Luke being selected both times in the next two class meetings is \(\frac{1}{1600}\).

Step by step solution

01

Calculate the probability of Luke being selected in the first class meeting

In order to calculate the probability of Luke being selected in the first class meeting, we can simply divide the number of favorable outcomes (1, as there is only one Luke) by the total number of possible outcomes (40, as there are a total of 40 students in the class). So, the probability of Luke being selected in the first class is: \(P(Luke \ selected \ in \ 1st \ class) = \frac{1}{40}\)
02

Calculate the probability of Luke being selected in the second class meeting

Similarly, for the second class meeting, we can calculate the probability of Luke being selected by dividing the number of favorable outcomes (1) by the total number of possible outcomes (40). So, the probability of Luke being selected in the second class is: \(P(Luke \ selected \ in \ 2nd \ class) = \frac{1}{40}\)
03

Calculate the probability of both events occurring

To find the probability of Luke being selected in both the first and second class meetings, we need to multiply the individual probabilities from Step 1 and Step 2. So, the probability of Luke being selected in both the first and second class meetings is: \(P(Luke \ selected \ both \ times) = P(Luke \ selected \ in \ 1st \ class) \times P(Luke \ selected \ in \ 2nd \ class) = \frac{1}{40} \times \frac{1}{40} = \frac{1}{1600}\) Hence, the probability of Luke being selected both times in the next two class meetings is \(\frac{1}{1600}\).

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Key Concepts

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

Random Selection
Random selection is a fundamental concept in probability that involves choosing individuals or events from a larger set to ensure each has an equal chance of being selected. In the context of our exercise, the professor selects a student at random to explain a homework problem. This randomness ensures that each student, including Luke, has an equal opportunity to be picked.

Randomness means that each of the 40 students has the same probability of being chosen during each class meeting. This can be visualized by thinking about drawing a name from a hat containing 40 evenly mixed names. Doing this without bias satisfies the condition of random selection.
  • Ensures fair selection.
  • Eliminates any bias in the process.
Favorable Outcomes
A favorable outcome in probability is the outcome we are interested in observing. In this problem, one favorable outcome is Luke being picked in one class session.

The total number of possible outcomes in a class meeting is 40, representing each student that could potentially be chosen. Hence, for Luke, only one specific outcome (his selection) is favorable out of these 40.
  • Favorable outcome = Luke gets selected.
  • Total possible outcomes = 40 students.
Understanding favorable outcomes allows us to set up the basic probability calculation by dividing the number of favorable outcomes by the total number of possible outcomes.
Independent Events
Independent events refer to occurrences where the outcome of one event does not affect the outcome of another. In our problem, Luke's selection in each class is treated as an independent event.

Being picked in the first and second class meetings are independent because selecting Luke the first time does not change his probability of being picked again during the second meeting. Each class meeting is a separate event with the same chance of choosing any student.
  • Outcome of one selection doesn't affect the other.
  • The randomness of selection ensures independence.
This characteristic is essential because it allows us to use the multiplication rule to find the probability of both events happening.
Multiplication Rule
The multiplication rule is used to calculate the probability of two independent events both happening. It states that the probability of both events occurring is the product of their individual probabilities.

As calculated in the exercise, the probability of Luke being selected the first time is \( \frac{1}{40} \), and the same for the second meeting. By the multiplication rule, you find the combined probability by multiplying these two probabilities:\[ P(Luke \ selected \ both \ times) = \frac{1}{40} \times \frac{1}{40} = \frac{1}{1600} \]
  • Used for independent events.
  • Probability of both events = Product of individual probabilities.
The multiplication rule simplifies handling compound events, making it easier to find the likelihood of occurrences like Luke being selected in two consecutive meetings.

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Most popular questions from this chapter

The paper "Predictors of Complementary Therapy Use Among Asthma Patients: Results of a Primary Care Survey" (Health and Social Care in the Community [2008]: \(155-164)\) described a study in which each person in a large sample of asthma patients responded to two questions: Question 1: Do conventional asthma medications usually help your symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? Suppose that this sample is representative of asthma patients. Consider the following events: \(E=\) event that the patient uses complementary therapies \(F=\) event that the patient reports conventional medications usually help The data from the sample were used to estimate the following probabilities: $$ P(E)=0.146 \quad P(F)=0.879 \quad P(E \cap F)=0.122 $$ a. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(E\) and not \(E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. The probability that an asthma patient responds that conventional medications do not help and that the patient uses complementary therapies. ii. The probability that an asthma patient responds that conventional medications do not help and that the patient does not use complementary therapies. iii. The probability that an asthma patient responds that conventional medications usually help or the patient uses complementary therapies. c. Are the events \(E\) and \(F\) independent? Explain.

The report "Improving Undergraduate Learning" (Social Science Research Council, 2011) summarizes data from a survey of several thousand college students. These students were thought to be representative of the population of all college students in the United States. When asked about an upcoming semester, \(68 \%\) said they would be taking a class that is reading-intensive (requires more than 40 pages of reading per week). Only \(50 \%\) said they would be taking a class that is writing-intensive (requires more than 20 pages of writing over the course of the semester). The percentage who said that they would be taking both a reading-intensive course and a writing-intensive course was \(42 \%\). a. Use the given information to set up a hypothetical 1000 table. b. Use the table to find the following probabilities: i. the probability that a randomly selected student would be taking at least one reading-intensive or writing-intensive course. ii. the probability that a randomly selected student would be taking a reading-intensive course or a writingintensive course, but not both. iii. the probability that a randomly selected student would be taking neither a reading-intensive nor a writing-intensive course.

False positive results are not uncommon with mammograms, a test used to screen for breast cancer. For a woman who has a positive mammogram, the probability that she actually has breast cancer is less than 0.05 if she is under 40 years old, and ranges from 0.050 to 0.109 if she is over 40 years old ("Breast Cancer Screenings: Does the Evidence Support the Recommendations?," Significance [August 2016]: 24-37). If a woman with a positive mammogram is selected at random, are the two events \(B=\) event that selected woman has breast cancer and \(A=\) event that selected woman is over 40 years old independent events? Justify your answer using the given information.

Four students must work together on a group project. They decide that each will take responsibility for a particular part of the project, as follows: Because of the way the tasks have been divided, one student must finish before the next student can begin work. To ensure that the project is completed on time, a time line is established, with a deadline for each team member. If any one of the team members is late, the timely completion of the project is jeopardized. Assume the following probabilities: 1\. The probability that Maria completes her part on time is 0.8 . 2\. If Maria completes her part on time, the probability that Alex completes on time is 0.9 , but if Maria is late, the probability that Alex completes on time is only \(0.6 .\) 3\. If Alex completes his part on time, the probability that Juan completes on time is 0.8 , but if Alex is late, the probability that Juan completes on time is only \(0.5 .\) 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(0.9,\) but if Juan is late, the probability that Jacob completes on time is only 0.7 . Use simulation (with at least 20 trials) to estimate the probability that the project is completed on time. Think carefully about this one. For example, you might use a random digit to represent each part of the project (four in all). For the first digit (Maria's part), \(1-8\) could represent on time, and 9 and 0 could represent late. Depending on what happened with Maria (late or on time), you would then look at the digit representing Alex's part. If Maria was on time, \(1-9\) would represent on time for Alex, but if Maria was late, only \(1-6\) would represent on time. The parts for Juan and Jacob could be handled similarly.

Consider a chance experiment that consists of selecting a customer at random from all people who purchased a car at a large car dealership during 2016 . a. In the context of this chance experiment, give an example of two events that would be mutually exclusive. b. In the context of this chance experiment, give an example of two events that would not be mutually exclusive.
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