CHANCE S. 



741 



Qkuet*. for the probability required, which, in iiumbcn, is 

 ' 3x25 



Cor. The probability of throwing one ace, and no 

 more, at one throw, by three dice, is also -^ '-. 



PROBLEM XIII. 



22. What is the probability of throwing one tee, 

 ami no more, in four throws ? 



SOLI-TION. This probability is the sum of the 

 pn liability of throwing an ace at the first throw, and 

 missmg it in all the remaining three throws ; and of 

 the probability of missing an ace at the first throw, 

 and throwing pne, and no more, in the three remain- 

 ing throws. 



riu- probability of throwing an ace at one throw 



is - ; and that of missing to throw an ace, in three 



/ 1^5 



successive throws, is , (Art. 16.) ; and there- 



n* 



fore the probability of throwing an ace at the first 

 throw, an<1 miss'ng it in the three following throws, is 



i.^^-J-rfC; 



n ' n j n 4 



Again, the probability of missing to throw an ace 



at one throw is , and that of throwing one ace, 



ft 



and no more, in three successive throws, is 



(Art. 21.); and therefore the probability of mis- 

 sing an ace at the first throw, and throwing one, 

 and no more, in the three following throws, is 



n w 3 n 4 



Therefore, the probability required is 

 (n I)* 3(" 5 _ JJM 1) 

 n 

 4- x 125 



4 

 125 



PROBLEM XIV. 



23. What is the probability of throwing one ace 

 and no more, iii m throws, m being any nurr* 

 whatever > 0> an( j 



SOLUTION. We have found in the Hth, 

 13th problems, that when the values of 

 and i, then the probabilities required -" 

 ' 



re 



, these problems, we 



the probability is 



1 



, 

 . , u 



T PC M !?'/' L By T 80 " 1 ^ 

 should find, that when i= 



^rdl*. Therefore, rending to the law of th6 



formation of these *""? J ia ea ^ tO ^ th j^ 

 in general, the pr^ab.l.ty of throwing one ace, and 



m ~ 

 no more, in 



Cor. The same, formula also expresses the proba- 

 bility of throwing a single ace at ODC throw with m 



dice. 



PliOBLEM XV. 



2k What is the probability of throwing two acei, 

 or more, in three throws, with a single die ? 



SOLUTION. Besides the event in question, there 

 are just other two that can possibly happen : either 

 never an ace will come up in the three throws ; or 

 (I ie one, and no more, will be thrown. These three 

 events then, making up certainty, the sum of their 

 probabilities must be unity'. Therefore, the proba- 

 bility re-quired will be found, by subtracting the other 

 two from 1. 



Now, the probability of throwing never an ace ra 



three throws is V^f- (Art. 16. );and the probabili- 



ty of throwing one ace, and no more, is 



(Art. 



23.); therefore, the probability of throwing two 

 aces, or more, in three throws, will be 

 l_Jj- l )L_ 9 ( *)' = n' ( P) 3(n I)' 



, 2161253x25 2 

 which, in numbers,. M 7^ = ofi* 



Cor. The probability of throwing two aces, or 

 more, at one throw, with three dice, is expressed by 

 the same formula. 



PROBLEM XVI. 



In numbers, 



Cor. The probability of throwing one ace, and 

 ao more, with four dice, at a single throw, is also 



jaces. 



25. What is the probability of throwin 

 or more, in four throws ? .a unity sub- 



SOLUTION. In this case we ""j^^g never aa 

 tract L-ir_2-, the probability . $ 



.il!^~-jL, the probabili- 

 ace in three throws, and^ j ^ ^^ 



ty of throwing one, i ) * n 4 --( I) 4 ; 



~~* JlSty required, which, in number*, ii 

 is the prr* 

 19 



14$r The same formula expresses the probability 

 . throwing two or more aces at one throw will 

 dice. 



PROBLEM XVII. 



26. What is the probability ef throwing two aces, 

 or mure, in m throw*, with a tingle die. 



SoLLT.ON.~By following the mode of reason n 

 empl.,yrd u. Fr b. 15. and 16. and observmg th 

 accord",,, to wuich th, nsults are formed.^ will 

 appear, that m being 



bnbility required is 



Cor. The probabilty of throwing two or more 

 aces, with m dice, at one throw, * expressed 

 same number. 



