Explanation Calculation: The data is given as learnigStats demonstration from McGraw-Hill’s Connect. It is also given that there are two riders on a city bus. For 2 riders: The probability of no two have the same birth day if there are 2 riders on a city bus represents the probability of no two riders have exact birth day. The probability of two riders have exact birth day is 0.0027 by using Birthday problem. The formula for finding the probability of no two riders have exact birth day is, P ( No two riders have exact birth day ) = 1 − P ( Two riders have exact birth day ) = 1 − 0.0027 = 0.9973 Thus, the probability that no two have the same birth day if there are 2 riders on a city bus is 0.9973. For 10 riders: The probability that no two have the same birth day if there are 10 riders on a city bus represents the probability of no ten riders have exact birth day. The probability of ten riders have exact birth day is 0.1169 by using Birthday problem. The formula for finding the probability of no ten riders have exact birth day is, P ( No ten riders have exact birth day ) = 1 − P ( Ten riders have exact birth day ) = 1 − 0.1169 = 0.8831 Thus, the probability that no two have the same birth day if there are 10 riders on a city bus is 0

Loose-leaf For Applied Statistics In Business And Economics

5th Edition

ISBN: 9781259328527

Author: David Doane, Lori Seward Senior Instructor of Operations Management

Publisher: McGraw-Hill Education

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Chapter 5, Problem 91CE

To determine

Find the probability that no two have the same birth day if there are 2 riders on a city bus. Also find it for 10, 20 and 50 riders.

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Chapter 5, Problem 90CE

Chapter 5, Problem 92CE

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Chapter 5 Solutions

Loose-leaf For Applied Statistics In Business And Economics

Ch. 5.1 - A credit card customer at Barnes and Noble can use...Ch. 5.1 - A survey asked tax accounting firms their business...Ch. 5.1 - A baseball player bats either left-handed (L) or...Ch. 5.1 - A die is thrown (1, 2, 3, 4, 5, 6) and a coin is...Ch. 5.2 - Prob. 5SE Ch. 5.2 - Prob. 6SE Ch. 5.2 - Prob. 7SE Ch. 5.2 - Prob. 8SE Ch. 5.2 - Prob. 9SE Ch. 5.2 - Prob. 10SE

Ch. 5.2 - Prob. 11SE Ch. 5.2 - Prob. 12SE Ch. 5.3 - Prob. 13SE Ch. 5.3 - Are these events collectively exhaustive or not?...Ch. 5.3 - Prob. 15SE Ch. 5.3 - Given P(A) = .70, P(B) = .30, and P(A B) = .00,...Ch. 5.3 - Prob. 17SE Ch. 5.3 - Prob. 18SE Ch. 5.3 - Prob. 19SE Ch. 5.3 - Prob. 20SE Ch. 5.3 - Prob. 21SE Ch. 5.3 - Prob. 22SE Ch. 5.4 - Given P(J) = .26, P(K) = .48. If A and B are...Ch. 5.4 - Given P(A) = .40, P(B) = .50. If A and B are...Ch. 5.4 - Prob. 25SE Ch. 5.4 - Prob. 26SE Ch. 5.4 - Prob. 27SE Ch. 5.4 - Prob. 28SE Ch. 5.4 - The probability that a student has a Visa card...Ch. 5.4 - Bob sets two alarm clocks (battery-powered) to be...Ch. 5.4 - A hospitals backup power system has three...Ch. 5.4 - Over 1,000 people try to climb Mt. Everest every...Ch. 5.4 - Suppose 50 percent of the customers at Pizza...Ch. 5.5 - Prob. 34SE Ch. 5.5 - The contingency table below summarizes a survey of...Ch. 5.5 - Prob. 36SE Ch. 5.5 - A survey of 156 introductory statistics students...Ch. 5.5 - This contingency table describes 200 business...Ch. 5.5 - Based on the previous problem, is major...Ch. 5.5 - The following contingency table shows average...Ch. 5.6 - Of grocery shoppers who have a shopping cart, 70...Ch. 5.6 - A study showed that 60 percent of The Wall Street...Ch. 5.7 - A drug test for athletes has a 5 percent false...Ch. 5.7 - Half of a set of the parts are manufactured by...Ch. 5.7 - An airport gamma ray luggage scanner coupled with...Ch. 5.8 - (a) Find 8! without a calculator. Show your work....Ch. 5.8 - Prob. 47SE Ch. 5.8 - In the Minnesota Northstar Cash Drawing, you pick...Ch. 5.8 - Prob. 49SE Ch. 5.8 - Prob. 50SE Ch. 5.8 - Until 2005, the UPC bar code had 12 digits (09)....Ch. 5.8 - Bob has to study for four final exams: accounting...Ch. 5.8 - Prob. 53SE Ch. 5.8 - Prob. 54SE Ch. 5.8 - Prob. 55SE Ch. 5.8 - Prob. 56SE Ch. 5 - Prob. 1CR Ch. 5 - Prob. 2CR Ch. 5 - Prob. 3CR Ch. 5 - Prob. 4CR Ch. 5 - Prob. 5CR Ch. 5 - Prob. 6CR Ch. 5 - Prob. 7CR Ch. 5 - In a contingency table, explain the concepts of...Ch. 5 - Prob. 9CR Ch. 5 - Prob. 10CR Ch. 5 - Prob. 11CR Ch. 5 - Prob. 60CE Ch. 5 - Prob. 61CE Ch. 5 - A judge concludes that there is a 20 percent...Ch. 5 - Prob. 63CE Ch. 5 - Prob. 64CE Ch. 5 - In the first year after its release, 83 percent of...Ch. 5 - Prob. 66CE Ch. 5 - Prob. 67CE Ch. 5 - If Punxsutawney Phil sees his shadow on February...Ch. 5 - Prob. 69CE Ch. 5 - Bob owns two stocks. There is an 80 percent...Ch. 5 - Prob. 71CE Ch. 5 - A study showed that trained police officers can...Ch. 5 - The probability that a 2011 Audi A8 will be stolen...Ch. 5 - The probability of being struck by lightning is...Ch. 5 - Prob. 75CE Ch. 5 - A certain model of remote-control Stanley garage...Ch. 5 - (a) In a certain state, license plates consist of...Ch. 5 - Prob. 78CE Ch. 5 - Prob. 79CE Ch. 5 - Prob. 80CE Ch. 5 - Prob. 81CE Ch. 5 - A certain airplane has two independent alternators...Ch. 5 - Prob. 83CE Ch. 5 - Prob. 84CE Ch. 5 - A turboprop aircraft has two attitude gyroscopes,...Ch. 5 - Which are likely to be independent events? For...Ch. 5 - In child-custody cases, about 70 percent of the...Ch. 5 - A web server hosting company advertises 99.999...Ch. 5 - Prob. 89CE Ch. 5 - The probability is 1 in 4,000,000 that a single...Ch. 5 - Prob. 91CE Ch. 5 - Prob. 92CE Ch. 5 - Four students divided the task of surveying the...Ch. 5 - Refer to the contingency table shown below. (a)...Ch. 5 - Prob. 95CE Ch. 5 - High levels of cockpit noise in an aircraft can...Ch. 5 - Prob. 97CE Ch. 5 - A biometric security device using fingerprints...Ch. 5 - Dolon Web Security Consultants requires all job...

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