THE THEORY OF PROBABILITIES 73 



If, however, at least 20 of the renuiining 2()8 subscribers initiate 

 their calls within the particular two minutes under consideration, 

 there will be no trunk immediately available for X. This follows 

 from assumptions B and C. 



Consider some one of these 268 other subscribers, for example Y. 

 The probability that Y calls in the two minutes under consideration is 

 by assumption A, the ratio of 2 minutes to 60 minutes, or 1/30, which 

 is exactly the same as the probability that he would throw an ace if 

 he were to make a single throw with a 30-face die. Likewise the 

 probability that still another subscriber calls in the two minutes 

 under consideration is exactly the same as the probability that this 



u One Hour <^ 



I I 



I I 



j }^_2Min.— J I 



I I I I 



II I 



I I \ y » 



I ! 



X^ I 



Fig. 2 



other subscriber should throw the ace in a single throw with a 30-face 

 die. 



It is evident then, that the probability that X fails to get a trunk 

 immediately is the same as the probability of throwing at least 20 aces 

 if 268 throws are made with a 30-face die. To facilitate the de- 

 termination of this probability and the solution of similar problems, 

 probability tables of a type shown in Table I have been computed.*^ 

 In the table, the average number of times an event may be expected 

 is represented by a. The probability that the event occurs at least 

 a greater number of times c = a -[- d \s represented by P. In the 

 problem under consideration, the average number of aces expected is 



269 

 8.96 = — — • Likewise in the present problem c = 20. Turning to 



Ok) 



the table, we find that corresponding to c = 20 and a = 8.96, the 

 value of the probability P is .OOL In the particular telephone prob- 

 lem under consideration this means that once in a thousand times 



* Table I is to be found in the Appendix and its origin is there explained. 



