ON DIRECT PROBABILITIES. 45 



in the preceding lottery, if I draw white I cannot draw 

 black, and vice versa, there being one drawing only 

 supposed. Hence the probability of drawing black or 

 white is T T ^ -f- T 5 - or i| . But suppose there had been 

 two lotteries, as follows : 



(7 white, 8 red) (5 black, 10 red), 



and I draw in one without knowing in which I am 

 drawing. The individual probabilities of white in the 

 two, if the lottery be known, are -$ and -f^ as before ; 

 but the question, what is the chance of drawing white 

 or black, and not red, is a different one from the pre- 

 ceding. If I draw without knowing in which lottery I 

 am drawing, my position is the same as if one lottery 

 had been thrown into the other, and I had drawn from 

 both in one, giving 



(7 white, 5 black, 18 red). 



This union, which will readily be admitted to make no 

 alteration in the probabilities, is not admissible unless 

 the total number of balls in both be the same. For, in 

 the mixture, the 8 red balls of the one and the 1 of 

 the other, are considered as severally of equal proba- 

 bility. But this they are not, unless the number of 

 possible cases in the two were the same. Suppose, for 

 instance, I had the following lotteries : 



(10 white, 10 red) (4 black, 5 red) ; 



the probability of each ball of the first is 1 TT , but of 

 each of the second -^. Before I can draw a white ball, 

 two events must happen. 1 . I must happen to select 

 the first lottery. 2. I must happen to draw a white 

 ball rather than a red. The chances of these are -i and 

 |-{{- or l ; consequently, \ X \ or -*- is my chance of a 

 white ball. But if I mix the two lotteries, that chance 

 will be I", which is more than J. The reason is, that I 

 have mixed together balls which had unequal chances of 

 being drawn, and treated them as if they were to have 

 equal chances. But it is evident, from the measure of 

 probability, that I do not alter the chances for an event 



