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A college job placement center has requests from five students for employment interviews. Three of these students are math majors, and the other two students are statistics majors. Unfortunately, the interviewer has time to talk to only two of the students. These two will be randomly selected from among the five. a. What is the sample space for the chance experiment of selecting two students at random? (Hint: You can think of the students as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D},\) and \(\mathrm{E}\). One possible selection of two students is \(\mathrm{A}\) and \(\mathrm{B}\). There are nine other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both selected students are statistics majors? d. What is the probability that both students are math majors? e. What is the probability that at least one of the students selected is a statistics major? f. What is the probability that the selected students have different majors?

Short Answer

Expert verified
The sample space for the chance experiment of selecting 2 students at random is {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}. The outcomes are equally likely, with each having a 1/10 chance. The probability of both students being statistics majors is 1/10, both being math majors is 3/10, at least one being a statistics major is 7/10, and the pair having different majors is 3/5.

Step by step solution

01

Sample Space

List all possible unique pairs of students without considering their majors. Label the students as A, B, C, D, and E. There are ten possible pairs: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
02

a. Sample Space

The sample space for the chance experiment of selecting two students at random is {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}.
03

b. Equally Likely Outcomes

Yes, the outcomes in the sample space are equally likely. Each pair has a 1/10 chance of being selected.
04

c. Probability of Both Statistics Majors

Let A and B be the statistics majors. We only have 1 pair consisting of both statistics majors, which is AB. As there are 10 possible pairs, the probability of selecting both statistics majors is 1/10.
05

d. Probability of Both Math Majors

Let C, D, and E be the math majors. We have 3 pairs consisting of only math majors: CD, CE, and DE. As there are 10 possible pairs, the probability of selecting both math majors is 3/10.
06

e. Probability of at Least One Statistics Major

To find the probability of selecting at least one statistics major, we can use the complement rule. The complement of selecting at least one statistics major is selecting no statistics majors, which means selecting both math majors. The probability of selecting both math majors is 3/10. Therefore, the probability of selecting at least one statistics major is 1 - 3/10 = 7/10.
07

f. Probability of Different Majors

To find the probability of selecting students with different majors, we can count the number of pairs with different majors. These pairs are AC, AD, AE, BC, BD, and BE. As there are 10 possible pairs, the probability of selecting students with different majors is 6/10 or 3/5.

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Key Concepts

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

Sample Space
In probability theory, the concept of a sample space is crucial. It refers to the set of all possible outcomes of a probability experiment. When you conduct a chance experiment, the sample space includes every possible result.
For the exercise given, the task is to choose 2 students out of 5, labeled as A, B, C, D, and E. Here, it shows all potential pairs that can be formed by selecting two students, without focusing on their individual majors.
So the sample space, which represents every unique pairing from the five students, is:
  • {AB, AC, AD, AE, BC, BD, BE, CD, CE, DE}
This list, with its ten pairs, represents all possible outcomes for the selection process. Understanding the sample space is the first step in analyzing any probability problem.
Equally Likely Outcomes
The idea of equally likely outcomes means that every outcome in a sample space has the same probability of occurring. In many problems, especially those involving fair selections or random draws, each outcome often carries the same weight.
For our current exercise, selecting pairs of students is done randomly, indicating that any pair is just as likely to be chosen as any other. This is confirmed by noting that each pair has a probability of \[\frac{1}{10}\]to be selected, which reflects equal likelihood since there are ten pairs total.
Understanding that outcomes are equally likely simplifies calculations and allows you to apply fundamental probability rules accurately.
Complement Rule
The complement rule is a valuable tool in probability theory. It allows you to find the probability of an event by considering the probability of its complement, or the event not happening. If you know the probability of an event, then its complement has a probability of one minus that event's probability.
In our exercise, to find the probability of selecting at least one statistics major, we use this rule. First, calculate the probability of the opposite event, which is selecting only math majors, and then subtract this probability from 1.
Given there are 3 pairs of math majors and 10 total pairs, the probability of choosing both math majors is \[\frac{3}{10}\]. Thus, the probability of selecting at least one statistics major is \[1 - \frac{3}{10} = \frac{7}{10}\].
This method is intuitive and often more straightforward than directly calculating the desired probability, especially when dealing with the "at least" type of things.

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Most popular questions from this chapter

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.

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?

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

a. Suppose events \(E\) and \(F\) are mutually exclusive with \(P(E)=0.14\) and \(P(F)=0.76\) i. \(\quad\) What is the value of \(P(E \cap F) ?\) ii. What is the value of \(P(E \cup F)\) ? b. Suppose that for events \(A\) and \(B, P(A)=0.24, P(B)=0.24\) and \(P(A \cup B)=0.48\). Are \(A\) and \(B\) mutually exclusive? How can you tell?

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