/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 The student council for a school... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The student council for a school of science and math has one representative from each of the five academic departments: biology (B), chemistry (C), mathematics (M), physics (P), and statistics (S). Two of these students are to be randomly selected for inclusion on a university-wide student committee (by placing five slips of paper in a bowl, mixing, and drawing out two of them). a. What are the 10 possible outcomes (simple events)? b. From the description of the selection process, all outcomes are equally likely. What is the probability of each simple event? c. What is the probability that one of the committee members is the statistics department representative? d. What is the probability that both committee members come from laboratory science departments?

Short Answer

Expert verified
a. The 10 possible outcomes are: BC, BM, BP, BS, CM, CP, CS, MP, MS, PS. b. The probability of each simple event is 0.1. c. The probability that one of the committee members is the statistics department representative is 0.4. d. The probability that both committee members come from laboratory science departments is 0.1.

Step by step solution

01

Calculation of Possible Outcomes (simple events)

The total possible outcomes of drawing 2 students from 5 can be calculated as combinations of 5 students taken 2 at a time. This can be represented mathematically as \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number and k is the number of items to select. In this case, \(C(5, 2) = \frac{5!}{2!(5-2)!} = 10\) possible outcomes. They are: BC, BM, BP, BS, CM, CP, CS, MP, MS, PS.
02

Probability of Each Simple Event

Since all outcomes are equally likely, the probability of each simple event occurring is the reciprocal of the total number of outcomes. This is calculated with the formula \(P(A) = \frac{1}{n}\), where P(A) is the probability of event A and n is the total number of outcomes. So, the probability of each simple event = \(\frac{1}{10}\).
03

Probability that a Committee Member is from the Statistics Department

We need to count combinations that include 'S'. From the list of 10 combinations, 4 of them (BS, CS, MS, PS) include 'S'. Using the formula in Step 2, the probability that one of the committee members is the statistics department representative is \( \frac{4}{10} = 0.4\).
04

Probability that both Committee Members are from Laboratory Science Departments

The laboratory science departments are Biology and Chemistry. Now, we are interested in the combinations that include only B and C. They are BC which is 1 of 10 outcomes. Hence, the probability that both committee members come from laboratory science departments is \(\frac{1}{10}=0.1\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

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?)

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four students \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2's book, 2 receives 4's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? Assuming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?

Suppose that we define the following events: \(C\) \(=\) event that a randomly selected driver is observed to be using a cell phone, \(A=\) event that a randomly selected driver is observed driving a passenger automobile, \(V=\) event that a randomly selected driver is observed driving a van or \(\mathrm{SUV}\), and \(T=\) event that a randomly selected driver is observed driving a pickup truck. Based on the article "Three Percent of Drivers on Hand-Held Cell Phones at Any Given Time" (San Luis Obispo Tribune, July 24,2001 ), the following probability estimates are reasonable: \(P(C)=.03\), \(P(C \mid A)=.026, P(C \mid V)=.048\), and \(P(C \mid T)=.019\) Explain why \(P(C)\) is not just the average of the three given conditional probabilities.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.