THE THF.ORY OF PROBABILITIES 77 



Case I 



Referring to Fig. 4 consider a group of n = 1486 lines, and let the 

 traffic be divided among the ten levels or directions available in 

 such a way that on the average V3 of the calls are made for a particular 

 direction or level. Let us suppose the circuit connections between 

 the line switches and first selectors to be such that the calls are dis- 

 tributed individually at random. By this is meant that the first 

 selector seized by a calling line is as likely to be one having access 

 to sub-group Gi as to sub-group G2, G3 or G4. Note carefully that 

 this distribution is assumed whether or not the calling line wants 

 the particular level under consideration. One way of securing this 

 random distribution by sub-groups would be to allow the line switches 

 first to choose by chance one of the four sub-groups of first selectors 

 and then to choose an idle first selector in the sub-group. 



Question — What is the probability that when subscriber X calls 

 he fails to obtain immediately a trunk to a second selector? It is 

 assumed that X obtained a first selector and that his call is for the 

 level under consideration. 



As before we are interested in the calls made during the two minutes 

 preceding the instant at which X calls. Let the number of these calls be 

 C. Of these C calls a certain number D want the level for which X 

 has called. If at least 10 of these D calls were distributed by the 

 line switches to first selectors having access to the same sub-group 

 as the one to which the first selector seized by X has access, then 

 there will be no idle trunk in the sub-group for X. Our telephone 

 trunking problem evidently transforms to the following series of 

 dice problems. 



1st. 1486 throws are made with a 30 face die giving C aces. 

 2nd. C throws are made with a 3 face die giving D aces. 

 3rd. D throws are made with a 4 face die giving x aces, and the 

 question is the probability that x is not less than 10. 



By the theory of dice (assuming no restriction ^ on the value of 

 C) the probability is the same as that of throwing at least 10 aces in 

 1486 throws with a die having 30 X 3 X 4 = 360 faces. The average 

 number of aces to be expected being 1486/360 = 4.13 the probability 

 tables give .01 as the answer. 



Case 2 



As in Case 1 assume that on the average 3^ of the calls are for 

 the level under consideration, but take n = 1725 for the number of 

 lines. Now suppose the circuit connections between line switches 



