Introduction to Probability and Its Applications, Third by Linda J. Young Richard L. Scheaffer

By Linda J. Young Richard L. Scheaffer

This article makes a speciality of the application of likelihood in fixing real-world difficulties for college students in a one-semester calculus-based chance path. idea is constructed to a realistic measure and level-headed in dialogue of its useful makes use of in fixing real-world difficulties. a number of purposes utilizing up to date actual facts in engineering and the existence, social, and actual sciences illustrate and encourage the various methods likelihood impacts our lives. The text's obtainable presentation rigorously progresses from regimen to more challenging difficulties to fit scholars of alternative backgrounds, and punctiliously explains how and the place to use tools. scholars occurring to extra complex classes in likelihood and information will achieve a pretty good heritage in primary strategies and idea, whereas scholars who needs to follow likelihood to their classes engineering and the sciences will advance a operating wisdom of the topic and appreciation of its sensible energy.

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Notice by looking at the Venn diagram that if we add P(A) 24 Chapter 2 Foundations of Probability and P(B), we have added the probability associated with A ∩ B twice and would need to subtract this probability to obtain P(A ∪ B); that is, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). 8. 8 ¯ Showing A ∪ B = A ∪ AB using Venn diagrams. A B A is shaded A B AB is shaded A B A B A ∪ AB is shaded A ∪ AB = A ∪ B B is shaded ¯ are disjoint sets, we have Now, because A and AB ¯ = P(A) + P(AB). ¯ P(A ∪ B) = P(A ∪ AB) ¯ where AB and AB ¯ are disjoint sets.

5? Notice by looking at the Venn diagram that if we add P(A) 24 Chapter 2 Foundations of Probability and P(B), we have added the probability associated with A ∩ B twice and would need to subtract this probability to obtain P(A ∪ B); that is, P(A ∪ B) = P(A) + P(B) − P(A ∩ B). 8. 8 ¯ Showing A ∪ B = A ∪ AB using Venn diagrams. A B A is shaded A B AB is shaded A B A B A ∪ AB is shaded A ∪ AB = A ∪ B B is shaded ¯ are disjoint sets, we have Now, because A and AB ¯ = P(A) + P(AB). ¯ P(A ∪ B) = P(A ∪ AB) ¯ where AB and AB ¯ are disjoint sets.

Solution Because three different account managers are to make the contacts, the choices are made without replacement. The order in which the account managers are assigned to the customers matters because the customers’ needs differ, and this is considered in making the assignments. Thus, the number of ways to assign 3 of the 12 account managers to the customers is the permutations of 3 objects selected from 12 objects, that is, 12! P312 = = 12 × 11 × 10 = 1320. 9! The director has 1320 possible assignments!

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