pectation, will give - X LJ , N for the probabili- 

 ty required. Hence it appears, that the probability 

 oft/ie happening <>/' tin> independent events is equal to 

 tin- product of the probabilities of their happening se- 



Cor. If the two events are both of the same kind, 

 then the probability will be -. .,. 



PnOBLEM VII. 



15. A person offers to lay a wager of L.I, that 

 out of a purse containing m-f-;i counters, of which 

 m are black, and n white, he will draw, blindfolded, 

 at the first trial, a black counter ; and also, that out 

 ot another purse containing M-f-N counters, of which 

 M are black, and N white, he will also draw at the 

 first trial a white counter ; and if either experiment 

 fail he ia to lose his wager. What is the probability 

 of his succeeding I 



SOLUTION. By reasoning as in the last problem, 

 it will appear, that the probability is expressed by 



: Xr? . VT : But we may also understand this 

 m -f- M + N 



problem to mean, that he is to fail of drawing a 

 white counter at the first trial, and to succeed in the 

 second : Now, the probability of drawing, a white 



counter the first trial being , that of failing will 



be 1 , (Art. 12.) ; and the probability of suc- 



N 

 ceeding in the second trial being v? ^M the product 



of these two fractions, or [I : ) XTT r^will 

 \ m -f n) M + N 



be the probability required, (Art. 14.) 



In like manner it appears, that the probability of 

 failing in both experiments is the product of the pro- 

 babilities of their failing separately, viz. 



CHANCES. 739 



JLw.. pening of the first two a a tingle erent, the proba. C!iar. 



bility of which is -r X~r (Art. 14.); therefore the 



Cor. If the events be both of the same kind, the 

 probability of failing in the first trial, and succeeding 



in the second, will be (l -jX , and that 



(71 \* 

 1 ) . 

 m + n/ 



16. We may deduce the probability of the hap- 

 pening of three independent events from that of two, 

 and the probability of four from that of three, and so 

 on, by reasoning as follows: Suppose that the re. 

 ceiving of a sum of money, as L.I, depends on the 

 happening of three events, the probabilities of which 



are -j- , , , respectively. Then, if the two first 



events had actually happened, so that the receiving 

 of the sum depended on the third event only, the 

 value of the expectation on that event would be 



prXL. 1, (Art. 12.) Therefore we may consider 



e 



~ X L. 1 as a sum to be received, provided the two 



first events happen. Now, we may regard the hap* 



value of the expectation on all the three is TrX-rX 

 -7-xL. 1, and the probability of their happening it 



In general, whatever be the number of independent 

 events, the probability of their alt happening it equal 

 io the continual product of their teparate probabiliiif*. 



PROBLEM VIII. 



17. What h the probability of throwing with a 

 single die one ace, or more, in two throwt, that it, 

 either at the first or second throw, or at both ? 



SOLUTION. In this problem, we may consider the 

 probability reqi ired as made up of two other proba- 

 bilities, viz. that of throwing an ace at the first throw, 

 and that of missing it at the first throw, and throw- 

 ing it at the second : Therefore, if we estimate these 

 separately, their sum will be the probability requi- 

 red. 



In order to render the conclusion general, let (T 

 (the number of faces of the die) =. Then the 

 probability of throwing an ace at the first throw is 



1 1 1 



; and that of missing it, 1 = . Also 



n n n 



the probability of throwing an ace at the second 

 throw is ; and, connecting this with the probabi- 

 lity of missing it the first trial, we liave the proba- 

 bility of missing it at the first trial and throwing 



it the second, expressed by X , (Art. 14.) 



Therefore the probability required is | 



gH1 _'_(!)*_ 11 



> -36* 



This problem may also be resolved by considering, 

 that the probability of missing an ace twice, together 

 with the probability of throwing an ace either at the 

 first or second throw, must amount to certaiuty, 

 which is measured by 1 : Therefore, if we estimate 

 the former of these probabilities, and subtract it 

 from unity, the remainder must be the latter. Now, 

 the probability of missing an ace at the first throw ia 



, (Art. 12.) and the probability of mining it 



the second is also -^- ; therefore the probability of 

 n 



missing it twice is ^ ;--; (Art. 14.) ; and hence 



the probability of throwing it once at least in two 

 , is 1- C=!di:, the same as be- 



fore. 



Cor. Hence it appears, that the probability of 

 throwing one ace or more at a single throw with two 



.. . 

 dice, u 



-.-^3-. For it JJ evidently the 



