/*! 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 29 A college student frequents one ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 29

A college student frequents one of two coffee houses on campus, choosing Starbucks \(70 \%\) of the time and Peet's \(30 \%\) of the time. Regardless of where she goes, she buys a cafe mocha on \(60 \%\) of her visits. a. The next time she goes into a coffee house on campus, what is the probability that she goes to Starbucks and orders a cafe mocha? b. Are the two events in part a independent? Explain. c. If she goes into a coffee house and orders a cafe mocha, what is the probability that she is at Peet's? d. What is the probability that she goes to Starbucks or orders a cafe mocha or both?

Short Answer

Expert verified
Answer: The probability that the student visits Starbucks or orders a cafe mocha, or both, is 0.88.

Step by step solution

01

a. Probability of visiting Starbucks and ordering a cafe mocha

To find the probability of the student visiting Starbucks and ordering a cafe mocha, we can use the conditional probability formula: P(S and M) = P(S) * P(M|S) In this case, P(S) = 0.7 and P(M|S) = 0.6. Therefore, we have: P(S and M) = 0.7 * 0.6 = 0.42
02

b. Are the two events independent?

To determine if the events are independent, we can check if P(S and M) = P(S) * P(M). First, we need to calculate P(M): - P(M) = P(M|S) * P(S) + P(M|P) * P(P) - P(M) = 0.6 * 0.7 + 0.6 * 0.3 - P(M) = 0.6 Now, let's check the independence condition: P(S and M) = 0.42 P(S) * P(M) = 0.7 * 0.6 = 0.42 Since P(S and M) = P(S) * P(M), the two events are independent.
03

c. Conditional probability of being at Peet's given a cafe mocha order

To find the probability that the student is at Peet's given that she ordered a cafe mocha, we can use the conditional probability formula: P(P|M) = P(P and M) / P(M) We already know P(M) = 0.6. We need to find P(P and M): P(P and M) = P(P) * P(M|P) P(P and M) = 0.3 * 0.6 = 0.18 Now we can calculate P(P|M): P(P|M) = P(P and M) / P(M) P(P|M) = 0.18 / 0.6 = 0.3
04

d. Probability of visiting Starbucks, ordering a cafe mocha, or both

To find the probability of visiting Starbucks or ordering a cafe mocha or both, we can use the formula for the probability of the union of two events: P(S or M) = P(S) + P(M) - P(S and M) We already know P(S) = 0.7, P(M) = 0.6, and P(S and M) = 0.42. Now we can calculate P(S or M): P(S or M) = 0.7 + 0.6 - 0.42 = 0.88

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 91影视!

Key Concepts

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

Independent Events
When we talk about independent events in probability, we're referring to scenarios where the outcome of one event does not affect the outcome of another. For instance, flipping a coin and rolling a die are independent events because the coin toss has no influence on the die roll.

To test for independence between two events, say A and B, we check whether the probability of both A and B occurring is the same as the product of their individual probabilities - that is, if P(A and B) = P(A) * P(B). If this condition is met, then A and B are independent. In our coffee house scenario, because the student chooses where to buy coffee and whether to buy a cafe mocha independently, we can say these two events are independent, which is demonstrated through the matching probabilities in step 2 of the solution.
Probability Formulas
The world of probability is rich with formulas to help us tackle a variety of questions. The fundamental one is the conditional probability formula, which tells us how likely an event is, given that another event has occurred. It's usually written as P(A|B), meaning the probability of A given B.

Another crucial one is the multiplication rule for independent events: if A and B are independent, then P(A and B) = P(A) * P(B). This is used in step 1 of our solution to find the probability of the student's coffee purchase behavior. Understanding how to apply these formulas is key to unraveling complex probability questions.
Union of Events
In probability theory, the union of events refers to occurrences where at least one of the events happens. The probability of the union of two events A and B (denoted as P(A or B)) can be calculated with the formula: P(A) + P(B) - P(A and B).

Why subtract P(A and B)? This part of the formula corrects for the fact that the intersection of A and B (when both events occur) has been counted twice. In the context of our coffee house example, we're interested in the chance that the student either goes to Starbucks, orders a cafe mocha, or does both, and this is where the formula comes into play in step 4. Understanding the union of events is essential in order to answer questions involving 'either/or' conditions, which are common in probability questions and real-world scenarios alike.

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

Suppose \(P(A)=.1\) and \(P(B)=.5 .\) $$\text { If } P(A \cup B)=.65, \text { are } A \text { and } B \text { mutually }$$$$\text { exclusive? }$$

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the probability that it gets by the first inspector is \(.1 .\) Of those that get past the first inspector, the second inspector will "miss" 5 out of \(10 .\) What fraction of the defective items get by both inspectors?

A student prepares for an exam by studying a list of 10 problems. She can solve 6 of them. For the exam, the instructor selects 5 questions at random from the list of \(10 .\) What is the probability that the student can solve all 5 problems on the exam?

Evaluate the permutations. $$ P_{1}^{20} $$

A study of drug offenders who have been treated for drug abuse suggests that the chance of conviction within a 2 -year period after treatment may depend on the offender's education. The proportions of the total number of cases that fall into four education/conviction categories are shown in the following table. Suppose a single offender is selected from the treatment program. Here are the events of interest: \(A:\) The offender has 10 or more years of education B: The offender is convicted within 2 years after completion of treatment Find the appropriate probabilities for these events: a. \(A\) b. \(B\) c. \(A \cap B\) d. \(A \cup B\) e. \(A^{c}\) f. \(A\) given that \(B\) has occurred g. \(B\) given that \(A\) has occurred
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation
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.