TAILIEUCHUNG - Introduction to Probability phần 4

Những con số còn lại được xác định bởi phương trình liên quan tái phát , đó là, nhập n cho 0 Trong trường hợp này, sự phân bố Benford có chức năng phân phối f (k) = log10 (k + 1) - log10 (k), 1 ≤ k ≤ 9. Mark Nigrini3 đã ủng hộ việc sử dụng phân phối Benford như một phương tiện thử nghiệm | . DISCRETE CONDITIONAL PROBABILITY 145 The sample space is R3 R X R X R with R 1 2 3 4 5 6 . If w 1 3 6 then X1 w 1 X2 w 3 and X3 w 6 indicating that the first roll was a 1 the second was a 3 and the third was a 6. The probability assigned to any sample point is m w 1 1 1 7 6 6 6 216 Example Consider next a Bernoulli trials process with probability p for success on each experiment. Let Xj w 1 if the jth outcome is success and Xj w 0 if it is a failure. Then X1 X2 . Xn is an independent trials process. Each Xj has the same distribution function mj where q 1 p. If Sn X1 X2 Xn then n P Sn j nip q j jj and Sn has as distribution the binomial distribution b n p j . Bayes Formula In our examples we have considered conditional probabilities of the following form Given the outcome of the second stage of a two-stage experiment find the probability for an outcome at the first stage. We have remarked that these probabilities are called Bayes probabilities. We return now to the calculation of more general Bayes probabilities. Suppose we have a set of events H1 H2 . Hm that are pairwise disjoint and such that the sample space Q satisfies the equation Q H1 u H2 U U Hm We call these events hypotheses. We also have an event E that gives us some information about which hypothesis is correct. We call this event evidence. Before we receive the evidence then we have a set of prior probabilities P H1 P H2 . P Hm for the hypotheses. If we know the correct hypothesis we know the probability for the evidence. That is we know P E Hj for all i. We want to find the probabilities for the hypotheses given the evidence. That is we want to find the conditional probabilities P Hj E . These probabilities are called the posterior probabilities. To find these probabilities we write them in the form PHE P Hi n E P Hi E P E E1 146 CHAPTER 4. CONDITIONAL PROBABILITY Disease Number having this disease The results - di 3215 2110 301 704 100 d2 2125 396 132 1187 410 d3 4660 510 3568 73 509 .

TỪ KHÓA LIÊN QUAN
Đã phát hiện trình chặn quảng cáo AdBlock
Trang web này phụ thuộc vào doanh thu từ số lần hiển thị quảng cáo để tồn tại. Vui lòng tắt trình chặn quảng cáo của bạn hoặc tạm dừng tính năng chặn quảng cáo cho trang web này.