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Chapter 6: Problem 25

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 0.6.

Step by step solution

01

Listing All Possible Outcomes

Identify all the possible combinations or outcomes of selecting 2 representatives from the 5 departments. Use the combination formula by which number of ways to choose 'k' from 'n' objects is denoted as \(C(n, k) = \frac{n!}{k!(n-k)!}\). For this case, \(C(5, 2) =10\), let's list out the 10 possible outcomes :\['BC', 'BM', 'BP', 'BS', 'CM', 'CP', 'CS', 'MP', 'MS', 'PS'\]
02

Determine The Probability Of Each Simple Event

Let A be the set of all possible outcomes then the probability of A happening is 1. The probability of each case is equal, hence each individual pair has a 1/10 chance of being chosen.
03

Calculating the probability of having a representative from the statistics department

Find the number of pairs that include the Statistics (S) department representative. From the list, you can see that 4 pairs contain 'S'. Hence the probability of drawing a Statistics representative is 4 out of 10, or 0.4.
04

Computing the Probability That Committee Members Come From Laboratory Science Departments

Here, the Laboratory Science departments are considered to be Biology (B) and Chemistry (C). Hence we're looking for pairs that includes either B or C, or both. The combinations which satisfy this are: 'BC', 'BM', 'BP', 'BS', 'CM', 'CP'. Therefore, the probability that both committee members come from laboratory science departments is 6/10 or 0.6.

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Key Concepts

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

Probability Theory
Probability Theory is an important branch of mathematics that deals with analyzing events governed by random processes. At its core, probability measures the likelihood that an event will occur. This is represented by a number between 0 and 1, where 0 indicates impossibility and 1 denotes certainty. In the context of our exercise, each simple event (choosing 2 students out of 5) has a probability attached to it, calculated by dividing 1 by the number of possible outcomes.

For instance, when we list all possible outcomes for choosing 2 students out of 5, we calculate that there are 10 possible combinations. Assuming each is equally likely, the probability of selecting any specific pair is:
  • Calculate the number of combinations: using the formula for combinations:
    \(C(n, k) = \frac{n!}{k!(n-k)!}\) gives \(C(5, 2) = 10\).
  • Probability of any specific outcome like 'BC' is: \(\frac{1}{10}\).
Understanding this foundational concept helps demystify how probability informs decision-making in uncertain situations.
Random Selection
Random Selection is a method where each item in a set has an equal chance of being chosen. It is a critical concept in creating unbiased samples in probability and statistics. In the context of the exercise, we utilize random selection to ensure that every student from the different departments has an equal opportunity to be part of the committee.

In the example provided, five slips of paper represent each department. These slips are mixed well before selecting two at random. This process is designed to eliminate any potential bias, ensuring that the chance of any particular pair being selected is purely driven by randomness.

This concept is crucial in various fields:
  • Sociology: For achieving unbiased survey results.
  • Science: In experiments to minimize allocation bias.
  • Business: For sampling customer feedback.
Essentially, random selection helps provide a fair representation that can be generalized to a larger population.
Statistical Methods
Statistical Methods are techniques used for collecting, analyzing, interpreting, and presenting data. They form the backbone of data-driven decision-making across numerous domains. In our exercise, several statistical methods were applied.

The first method was listing possible outcomes using combinations, which is a statistical tool for counting the ways items can be selected without considering order.

Another method involves calculating the probability of certain conditions, such as drawing a representative from the statistics department. Here, you count the favorable outcomes ('BS', 'CS', 'MS', 'PS') and divide by the total number of outcomes:
  • Favorable outcomes involving 'S': 4
  • Total possible outcomes: 10
    Thus, probability = \(\frac{4}{10} = 0.4\)
Lastly, these methods also allow us to calculate probabilities for specific conditions, like choosing representatives from laboratory sciences. This involves identifying combinations ('BC', 'BM', 'BP', 'BS', 'CM', 'CP') and computing probabilities. Statistical methods thereby offer a structured way to address complex random processes, making them easier to understand and predict.

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Most popular questions from this chapter

A theater complex is currently showing four R-rated movies, three \(\mathrm{PG}-13\) movies, two \(\mathrm{PG}\) movies, and one \(\mathrm{G}\) movie. The following table gives the number of people at the first showing of each movie on a certain Saturday: $$ \begin{array}{rlc} \text { Theater } & \text { Rating } & \begin{array}{l} \text { Number of } \\ \text { Viewers } \end{array} \\ \hline 1 & \mathrm{R} & 600 \\ 2 & \mathrm{PG}-13 & 420 \\ 3 & \mathrm{PG}-13 & 323 \\ 4 & \mathrm{R} & 196 \\ 5 & \mathrm{G} & 254 \\ 6 & \mathrm{PG} & 179 \\ 7 & \mathrm{PG}-13 & 114 \\ 8 & \mathrm{R} & 205 \\ 9 & \mathrm{R} & 139 \\ 10 & \mathrm{PG} & 87 \\ \hline \end{array} $$Suppose that a single one of these viewers is randomly selected. a. What is the probability that the selected individual saw a PG movie? b. What is the probability that the selected individual saw a PG or a PG-13 movie? c. What is the probability that the selected individual did not see an R movie?

Suppose that a six-sided die is "loaded" so that any particular even-numbered face is twice as likely to be observed as any particular odd-numbered face. a. What are the probabilities of the six simple events? (Hint: Denote these events by \(O_{1}, \ldots, O_{6}\). Then \(P\left(O_{1}\right)=p\), \(P\left(O_{2}\right)=2 p, P\left(O_{3}\right)=p, \ldots, P\left(O_{6}\right)=2 p .\) Now use a condi- tion on the sum of these probabilities to determine \(p .\) ) b. What is the probability that the number showing is an odd number? at most \(3 ?\) c. Now suppose that the die is loaded so that the probability of any particular simple event is proportional to the number showing on the corresponding upturned face; that is, \(P\left(O_{1}\right)=c, P\left(O_{2}\right)=2 c, \ldots, P\left(O_{6}\right)=6 c .\) What are the probabilities of the six simple events? Calculate the probabilities of Part (b) for this die.

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: a. The probability that grip size is \(4 \frac{1}{2}\) in. (event \(\mathrm{A}\) ) is $$ P(A)=P\left(O_{2} \text { or } O_{5}\right)=.20+.15=.35 $$ How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) in. c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) in.?

N.Y. Lottery Numbers Come Up 9-1-1 on 9/11" was the headline of an article that appeared in the San Francisco Chronicle (September 13,2002 ). More than 5600 people had selected the sequence \(9-1-1\) on that date, many more than is typical for that sequence. A professor at the University of Buffalo is quoted as saying, "I'm a bit surprised, but I wouldn't characterize it as bizarre. It's randomness. Every number has the same chance of coming up." a. The New York state lottery uses balls numbered \(0-9\) circulating in three separate bins. To select the winning sequence, one ball is chosen at random from each bin. What is the probability that the sequence \(9-1-1\) is the sequence selected on any particular day? (Hint: It may be helpful to think about the chosen sequence as a threedigit number.) b. What approach (classical, relative frequency, or subjective) did you use to obtain the probability in Part (a)? Explain.

The general addition rule for three events states that $$ \begin{aligned} P(A \text { or } B \text { or } C)=& P(A)+P(B)+P(C) \\ &-P(A \text { and } B)-P(A \text { and } C) \\ &-P(B \text { and } C)+P(A \text { and } B \text { and } C) \end{aligned} $$ A new magazine publishes columns entitled "Art" (A), "Books" (B), and "Cinema" (C). Suppose that \(14 \%\) of all subscribers read A, \(23 \%\) read \(\mathrm{B}, 37 \%\) read \(\mathrm{C}, 8 \%\) read \(\mathrm{A}\) and \(\mathrm{B}, 9 \%\) read \(\mathrm{A}\) and \(\mathrm{C}, 13 \%\) read \(\mathrm{B}\) and \(\mathrm{C}\), and \(5 \%\) read all three columns. What is the probability that a randomly selected subscriber reads at least one of these three columns?
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