By Hall E. H.

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Assuming we have k objects and n boxes, collocations of this type are in a one-to-one correspondence with the class of surjective maps Snk from {1, . . , k} onto {1, . . , n}, thus there are n Snk = (−1)j j =0 n (n − j )k j collocations of k pairwise different into n pairwise different boxes that place at least one object in each box. Another way to compute the previous number is the following. Assume i1 , . . , in objects are located in the boxes 1, . . e. i1 + · · · + in = k and i1 , . .

7; for the moment, let us think of E as a possible proper class of subsets of with the properties (i)–(iv) above. A family E of events of satisfying properties (i), (ii) and (iii) above is called an algebra of subsets of . e. ∞ (v) For any sequence Ai ⊂ E we have ∪∞ i=1 Ai ∈ E and ∩i=1 Ai ∈ E. This property will bring many further properties that will be shown later. Clearly this property boils down to (iii) when E is a ﬁnite family. Moreover, by De Moivre formulas it can be also simpliﬁed to: (vi) If (ii) holds, then for any sequence Ai ⊂ E either ∪∞ i=1 Ai ∈ E or ∩∞ i=1 Ai ∈ E.

E. P({x}) = (1 − p)k−1 p. 38 Compute the probability of getting k failures before obtaining the ﬁrst success in a Bernoulli process of trials. Solution. Let p be the probability of success in a single trial, 0 ≤ p ≤ 1. It sufﬁces to play k + 1 trials and compute the probability of the (k + 1)-tuple x = (0, 0, 0, . . , 0, 1). Therefore, P({x}) = (1 − p)k p. 39 Compute the probability of getting k failures before obtaining the nth success in a Bernoulli process of trials. Solution. Let p be the probability of success in a single trial, 0 ≤ p ≤ 1.