By L. Sucheston

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Additional info for Contributions to Ergodic Theory and Probability

Example text

Each such event therefore requires the occurrence of n definite results, of k occurrences and (n – k) failures of the event A. By the multiplication rule the probability of each such event is 31 pk(1 – p)n–k and the number of them is equal to Cnk , to the number of n elements taken k at a time. (n − k + 1) k! for determining the probability sought. (n − k + 1) k p (1 –p)n–k. k! 3) It is often expedient to express Cnk in a somewhat different way. Multiply its numerator and nominator by (n – k)(n – k – 1) … 2·1.

In the following table we entered the probabilities of each result calculated by the multiplication rule for independent events. The numbers of points gained by the shots are denoted, respectively, by ξ and η. ] The table shows that the sum ξ + η takes values 3, 4, 5 and 6. Value 2 is impossible since its probability is zero25. 04. The arrival of one of the following results […] is necessary and sufficient for ξ + η = 4. 24. (III) The sum of the probabilities is unity. Each law of distribution ought to possess this property since we deal here with the sum of the probabilities of all possible values of a random variable; that is, with the sum of the probabilities of some complete group of events.

Denote their probabilities by p1, p2 and p3, so that p3, for example, corresponds to hitting region I. The possible values of the random variable under consideration are the same for all shots but their probabilities can essentially differ. Such differences obviously determine the differences between the skills of the shots. 6 respectively. If a shot fires 12 times, the possible numbers of hit-points occurring in each region are 0, 1, 2, …, 11, 12. By itself, this information does not yet allow us to judge his skill.