742 



H A N C E S. 



Chances. 



PROBLEM XVIII. 



27. The probability of throwing two aces, and no 

 more, in three throws, is required. 



SOLUTION. The probability of throwing two aces 

 or more, in three throws, is manifestly made up of 

 the probability of throwing two aces and no more, 

 and of the probability of throwing three aces ; there- 

 fore, if from the first of these we subtract the third, 

 the remainder must be the second. 



The probability of throwing two aces or more, in 



. 3 ( l)^ 3(n 1 )* . 

 three throws, is ^ * (Art. 24.) 



and the probability of throwing three aces is - ; 



therefore the probability of throwing two aces, and 

 HO more, in three throws, is 



3 - ti 3 "" ri* 



which, in numbers, is . 

 7 



Cor. The same formula expresses the probability 

 of- throwing two aces, and no more, with three dice, 

 at one throw. 



PROBLEM XIX. 



28. The probability of throwing two aces, and no 

 more, in four throws, is required. 



SOLUTION. This probability may be considered 

 5 composed of the probability of bringing up an ace 

 rtfirst throw, and only one ace more in the three 

 an avng throws ; and of the probability of missing 

 no more, first throw, and throwing two aces, and 

 The prohe remaining throws. 



i *i ; and that S? f throwin S an ace the fif 8t throw 

 wing one ace, and no more, in the 

 three following, is-^-i \ 



th*fi,.ci r *v wi '- (Art. 21.); therefore, 



tfte first of the two probat/. 



quired is 1 gO 1 *)* __ 3('* composing that re- 



n ' 3 tf?* 



The probability of missing an ace t*. 

 n 1 . _ , .,. r . * 



J and that of. throwing two aces, 



in the remaining throws, is ^T!lU (Art. 2, 



therefore, the second of the two probabilities is 

 (nl) ^ 3(n l)_3(nlY 



n #4 



Therefore the probability required is 



-st throw is 



more 



3(n~lJ 3(n l)*_6(fl 



- , m ^ T 



which, in numbers, is . 

 216 



Cor. The probability of throwing two aces and 

 no. more at one throw with four dice, is expressed by 

 the same formula. 



PROBLEM XX. 



Proceeding as in the last problem, the probability Chances, 

 of throwing an ace the first throw, and only one more > T"""'' 

 in the four remaining throws, will be, by Art. 12 and 

 22 



L 4( 1)?_ 4(n 1)J 

 n ' n 4 n s 



And the probability of missing an ace the first throw, 

 and throwing two and no more in the remaining 

 throws, will be, by Art. 12 and 28, 



nl 6(nl)*_ 6 (nl)* 

 n n 4 n 5 



Therefore, the probability of throwing two aces and 

 no more in five throws, is 





6( K 1) } _ 10( 1) 



** 



which, expressed in numbers, is ** -. 



Cor. This formula expresses also the probability 

 of throwing two aces at one throw of five dice. 



PROBLEM XXI. 



30. The probability of throwing two aces, and no 

 more, in m throws, is required. 



It appears from the solution of the three last pro- 

 blems, that when the values of m are 3, 4, and 5, the 

 probabilities in question are, 



3.2 1 4.3 (nl) 1 - 5.(nV) 



h2 ~^~ > 1^ ~~*~ ' a L2""^ ' 

 respectively. Hence, by induction, we may infer, 

 that m being any number whatever, the probability 

 required is, 



m(m 1) (n l) m -* 



T.2 



Cor. This formula also expresses the probability 

 of throwing two aces and no more at one throw with- 

 m dice. 



PROBLEM XXII. 



31. The respective probabilities of throwing any 

 number of points with two dice, are required 



SOLUTION. It will appear upon consideration, 

 that 2 points and 12 points can each be thrown in one 

 way only, the first by the two aces coming up, and 

 the second by the two sixes. 

 1 ma each 



. 

 each be thrown in two ways, 



, 



i -'s, and 7 in six ways. Hence th 

 ATd t'he Dances is 2+4+6 + 8+ 10+6= 

 h Pbilof *" ?* "* P " 



The probabiii 



The probability ,: , 



The probability o* - 



The probability of\ 



And the probability o?^ in 7 j, ^^ 



PROBLEM Xx^x. 



out 



; 8 the proba- 



