By Sheldon M. Ross

**A First path in likelihood, 8th Edition**, gains transparent and intuitive reasons of the math of likelihood concept, impressive challenge units, and a number of diversified examples and functions. This ebook is perfect for an upper-level undergraduate or graduate point creation to chance for math, technology, engineering and enterprise scholars. It assumes a history in uncomplicated calculus.

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**Extra resources for A First Course in Probability (8th Edition)**

**Example text**

Summary 15 But, on letting y1 = x1 + 1, yi = xi , i = 2, . . , m, ym+1 = xm+1 + 1, we see that this number is equal to the number of positive vectors (y1 , . . 1, there are such outcomes, in agreement with the results of Example 4c. Suppose now that we are interested in the number of outcomes in which each pair of defective items is separated by at least 2 functional items. By the same reasoning as that applied previously, this would equal the number of vectors satisfying the equation x1 + · · · + xm+1 = n − m x1 Ú 0, xm+1 Ú 0, xi Ú 2, i = 2, .

Finally, since each of the n married couples can be arranged next to each other in one of two possible ways, it follows that there are 2n (20 − n − 1)! arrangements that result in a speciﬁed set of n men each sitting next to their wives. Therefore, P(Ei1 Ei2 · · · Ein ) = 2n (19 − n)! (19)! 4, we obtain that the probability that at least one married couple sits together, namely, 10 1 21 (18)! − (19)! 10 2 22 (17)! + (19)! 10 3 23 (16)! − ··· − (19)! 3395. ∗ EXAMPLE 10 10 210 9! 6605 (19)! 5o Runs Consider an athletic team that had just ﬁnished its season with a ﬁnal record of n wins and m losses.

5 SAMPLE SPACES HAVING EQUALLY LIKELY OUTCOMES In many experiments, it is natural to assume that all outcomes in the sample space are equally likely to occur. That is, consider an experiment whose sample space S is a ﬁnite set, say, S = {1, 2, . . , N}. ), that P({i}) = 1 N i = 1, 2, . . , N 34 Chapter 2 Axioms of Probability From this equation, it follows from Axiom 3 that, for any event E, P(E) = number of outcomes in E number of outcomes in S In words, if we assume that all outcomes of an experiment are equally likely to occur, then the probability of any event E equals the proportion of outcomes in the sample space that are contained in E.