By William Feller

ISBN-10: 0471257095

ISBN-13: 9780471257097

Significant adjustments during this version comprise the substitution of probabilistic arguments for combinatorial artifices, and the addition of latest sections on branching procedures, Markov chains, and the De Moivre-Laplace theorem.

**Read Online or Download An Introduction to Probability Theory and Its Applications, Volume 2 PDF**

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**Extra resources for An Introduction to Probability Theory and Its Applications, Volume 2**

**Example text**

Our probability is a weighted average of the probability in the two different cases, flood-damaged or not, and the weights are the corresponding probabilities of the cases. The example illustrates the idea behind the following important result. 1 (Law of Total Probability). Let B1 , B2 , . . be a sequence of events such that (a) P(Bk ) > 0 for k = 1, 2, . . (b) Bi and Bj are disjoint whenever i = / j ∞ (c) S = Bk k=1 Then, for any event A, we have ∞ P(A) = P(A|Bk )P(Bk ) k=1 Condition (a) is a technical requirement to make sure that the conditional probabilities are defined, and you may recall that a collection of sets satisfying (b) and (c) is called a partition of S.

Let us first solve this by using the definition. We have P(A) = 1 P(A ∩ H) = P(ace of hearts) = 52 and hence 4 52 , P(H) = 41 , and P(A ∩ H) = P(A)P(H) so that A and H are independent. Intuitively, the events give no information about 4 1 = 13 and if we are given the each other. The probability of drawing an ace is 52 1 information that the chosen card is a heart, the probability of an ace is still 13 . The proportion of aces is the same in the deck as within the suit of hearts. 36. Consider the previous example but suppose that we have removed the 2 of spades from the deck.

This example indicates that in order to call three events independent, we want each event to be independent of any combination of the other two. It turns out that the following definition guarantees this (see Problem 53). 4 Ironically, in the first edition of his excellent book Innumeracy: Mathematical Illiteracy and Its Consequences, John Allen Paulos described this problem a bit obscurely [4]. His terse formulation was “Consider now some randomly selected family of four. ” and he went on to claim that the probability is 23 .

### An Introduction to Probability Theory and Its Applications, Volume 2 by William Feller

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