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Chapter 6: Problem 57

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. a. Suppose that the hosts randomly select two answers from those submitted with replacement. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

Short Answer

Expert verified
The probability that both answers are correct with replacement is \(\frac{1}{256}\) or \(0.00391\) and the probability without replacement is \(\frac{49}{12784}\) or \(0.00383\).

Step by step solution

01

Calculate the Probability with Replacement

The total number of answers submitted is 800, and the number of correct answers submitted is 50. The probability of selecting a correct answer on a given attempt - \(P(Correct)\) - is therefore \(\frac{50}{800}\), which simplifies to \(\frac{1}{16}\). When replacement is involved, the probabilities don't change with each attempt because the total number of answers remains the same. Therefore, the probability of getting two correct answers when picking with replacement is the product of the probabilities of picking a correct answer twice \((\frac{1}{16})^2\), which equals \(\frac{1}{256}\).
02

Calculate the Probability without Replacement

When picking without replacement, the pool of answers decreases by one each time an answer is selected. The probability of picking a correct answer on the first attempt - \(P(Correct1)\) - remains \(\frac{50}{800}\) or \(\frac{1}{16}\). However, the probability of picking a correct answer on the second attempt - \(P(Correct2 | Correct1)\) - is \(\frac{49}{799}\), as both the numerator and the denominator decrease by one. The probability of picking two correct answers without replacement is the product of these probabilities, \(\frac{1}{16} * \frac{49}{799} = \frac{49}{12784}\).
03

Compare the Two Probabilities

By calculating the decimal equivalent of both probabilities, we find that the probability with replacement is \(0.00391\) while the probability without replacement is \(0.00383\). Though they are close, it is slightly more likely to get two correct answers when replacing after each pick than when not replacing.

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Key Concepts

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

Conditional Probability
Conditional probability helps us determine the likelihood of an event occurring given a certain condition. In the car question, if the first answer selected is correct, the probability of the second answer also being correct changes depending on whether or not the first answer is replaced.
When dealing with events without replacement, the probability of the second event is dependent on the outcome of the first event. For instance, if you've already picked a correct answer once and do not replace it, there is one less correct answer available in the pool. This is why conditional probability is so essential to understand in these scenarios.
  • Formula for conditional probability is: \( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
  • "\(P(A|B)\)" reads as the probability of event A given event B has occurred.
  • It adjusts the sample space based on prior outcomes.
Replacement vs. Non-Replacement
Replacement and non-replacement are critical concepts when dealing with probabilities, as they define whether the chosen item is placed back into the sample pool after being selected.
  • With Replacement: The probability for each selection remains constant because the sample pool does not change. For example, if you select a correct answer and replace it, the next choice has the same probability of being correct.
  • Without Replacement: The sample pool changes with each selection, meaning that the probabilities for subsequent selections are altered. When you pick a correct answer and do not replace it, there is one less correct answer available, decreasing the probability of picking another correct answer.

In our exercise, this distinction affects how we calculate the likelihood of getting two correct answers when choosing with or without replacement.
Calculating Probabilities
To calculate probabilities, we need to determine how likely an event is to happen relative to the total number of possible outcomes. This involves simple division of favorable outcomes by the total outcomes.
For example, in the exercise, we have 800 total answers and 50 correct answers. Thus, the probability of picking a correct answer on your first try is \( \frac{50}{800} = \frac{1}{16} \).
  • With Replacement: The probability for both selections is the product of picking a correct answer each time, i.e., \( \left(\frac{1}{16}\right)^2 = \frac{1}{256} \).
  • Without Replacement: The probability changes after the first pick. If the first answer is correct, only 49 correct answers remain out of 799 total, making the probability for the second correct pick \( \frac{49}{799} \). The combined probability then is \( \frac{1}{16} \times \frac{49}{799} = \frac{49}{12784} \).

Taking the decimal form helps in comparison, where we found the probability of both correct picks with replacement as 0.00391 and without replacement as 0.00383. This subtle difference is important for understanding how these two concepts affect outcome probabilities.

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Most popular questions from this chapter

A large cable company reports the following: \- \(80 \%\) of its customers subscribe to cable TV service \- \(42 \%\) of its customers subscribe to Internet service \- \(32 \%\) of its customers subscribe to telephone service \- \(25 \%\) of its customers subscribe to both cable TV and Internet service \- \(21 \%\) of its customers subscribe to both cable TV and phone service \- \(23 \%\) of its customers subscribe to both Internet and phone service \- \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. Find and interpret the following probabilities: a. \(P(\) cable TV only \()\) b. \(P\) (Internet \(\mid\) cable TV) c. \(P\) (exactly two services) d. \(P\) (Internet and cable TV only)

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) -belongs to a medical clinic that always has a physician at each of stations 1,2, and \(3 .\) During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1, \mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station \(1 .\) a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1, then the nine in which \(\mathrm{P}_{1}\) goes to station 2, and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station 2 . e. Identify outcomes in each of the following events: \(B^{C}, C^{C}, A \cup B, A \cap B, A \cap C\).

There are two traffic lights on the route used by a certain individual to go from home to work. Let \(E\) denote the event that the individual must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=.4, P(F)=.3\), and \(P(E \cap F)=.15\) a. What is the probability that the individual must stop at at least one light; that is, what is the probability of the event \(E \cup F\) ? b. What is the probability that the individual needn't stop at either light? c. What is the probability that the individual must stop at exactly one of the two lights? d. What is the probability that the individual must stop just at the first light? (Hint: How is the probability of this event related to \(P(E)\) and \(P(E \cap F)\) ? A Venn diagram might help.)

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 schedule 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 \(.8\). 2\. If Maria completes her part on time, the probability that Alex completes on time is \(.9\), but if Maria is late, the probability that Alex completes on time is only \(.6 .\) 3\. If Alex completes his part on time, the probability that Juan completes on time is \(.8\), but if Alex is late, the probability that Juan completes on time is only \(.5 .\) 4\. If Juan completes his part on time, the probability that Jacob completes on time is \(.9\), but if Juan is late, the probability that Jacob completes on time is only .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.

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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