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By Whitehead J., Zhou Y., Patterson S.

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Extra info for [Article] Easy-to-implement Bayesian methods for dose-escalation studies in healthy volunteers

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Hence, the number of possible combinations, is given by n k = n! k! (n − k)! 28. The number of combinations of two out of the four letters A, B, C, and D is found by letting n = 4 and k = 2. It is 4 2 = 4! = 6, 2! 2! consistently with the listing given earlier. It is worth observing that counting arguments sometimes lead to formulas that are rather difficult to derive algebraically. One example is the binomial formula n n k p (1 − p)n−k = 1 k k=0 46 Sample Space and Probability Chap. 5. Here is another example.

Bayes’ Rule Let A1 , A2 , . . , An be disjoint events that form a partition of the sample space, and assume that P(Ai ) > 0, for all i. Then, for any event B such that P(B) > 0, we have P(Ai )P(B | Ai ) P(B) P(Ai )P(B | Ai ) = . P(A1 )P(B | A1 ) + · · · + P(An )P(B | An ) P(Ai | B) = To verify Bayes’ rule, note that P(Ai )P(B | Ai ) and P(Ai | B)P(B) are equal, because they are both equal to P(Ai ∩ B). This yields the first equality. The second equality follows from the first by using the total probability theorem to rewrite P(B).

The theorem is visualized and proved in Fig. 12. Intuitively, we are partitioning the sample space into a number of scenarios (events) Ai . Then, the probability that B occurs is a weighted average of its conditional probability under each scenario, where each scenario is weighted according to its (unconditional) probability. One of the uses of the theorem is to compute the probability of various events B for which the conditional probabilities P(B | Ai ) are known or easy to derive. The key is to choose appropriately the partition A1 , .

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