740 CHANCES. 



whether we throw twice successively with one die, or probability of throwing- aa.ac*, or more, with tn dice Ghancei. 

 once with two dice. at one throw. N -*- k y-** 1 . 



PROBLEM IX. 



18. What is the probability of throwing an ace, 

 or any proposed face of the die, in three throws, 

 that ia, either at the first, second, or third throw ? 



SOLUTION. The probability required may be con- 

 sidered as composed of the separate probabilities of 

 two events, viz. Jirst, of the probability of throw- 

 ing an ace either at the first or second throw ; and, 

 secondly, of the probability of missing it both at the 

 -first and second throws, and throwing it at the third. 

 The probability of throwing an ace in two 

 throws ha* been found, by the last problem, to be 



n l^Z-Z- . aru l therefore (Art. 14.) the proba- 

 n* 



n * f n i\ 



bility of missing it both throws, is 1 ^-j 



:::- ~. Again, the probability of throwing an 



II 



1 



-ace at the third throw, is - ; and consequently the 



ft 



.probability of missing it the first two throws, and 



1 f n IV (!)* 

 succeeding at the third, is - . -^ * = i '. 



ft 91 W 



Adding now the two probabilities, we have the pro- 

 'bability required, expressed by 



n 1 ri* w j 



.In numbers the probability is -?^- s . 



The problem may also be resolved, by finding the 

 probability of missing an ace three times successively, 



... .... n 1 n 1 n 1 



which will be . . (Art. ;6. ), and 



n -n n 



.this subtracted from unity, gives us 1 - '- 



for the probability of throwing an ace 



pnce or more in three throws, as before, 



Cor^ Hence it appears, that the probability of 

 throwing one ace, at least, by three dice, at one single 



. 3_(_i)3 

 throw, is - ^ - ', 

 n* 



PROBLEM X. 



19. What is the probability of throwing one ace 

 or more in m throws, wj being put for any number 

 whatever ? 



SQLVTJoN.Following the secc-nd mode of solu- 

 tion employed in the two preceding problems, we 

 find tha.t the probability of missing an ace m. times 

 successively, is (by Art. 16..) 



, 

 and this expression subtracted frpm unity, leaves 



/or the probability of throwing one ace, or more, in 

 fn successive throws, 



0r.The iwe expression is the measure of the 



PROBLEM XT. 



20. What is the probability of throwing one ace, 

 and no more, with a single die, in two throws ? 



SOLUTION. The probability of throwing one ace, 

 or more, at two throws, may evidently be regarded 

 as the sum of two probabilities, viz. 



1st, The probability of throwing one ace and no 

 more. 



2dly, The probability of throwing two aces. 



Therefore, the probability of throwing one ace, 

 and no more, must be measured by the difference be- 

 tween the probability of throwing one ace or more 

 and the probability of throwing two aces at two 

 throws. 



Now, the first of the two latter probabilities is 



u i /,, i u 



V L I 



(Art. 17.) and the second is - a (Art. 



13.) and their difference, or the probability required, 

 n*(niy 1 2(721) 5 



| C v / , , , ^ '_ 



Cor. The probability of throwing one ace, and 

 no more, at a single throw, with two dice, is also 



PROBLEM XII. 



21. What is the probability of throwing one ace, 

 and no more, in three throws ? 



SOLUTION. This probability may be considered 

 as made up of two probabilities, viz. that of throw- 

 ing an ace at the first throw, and missing an ace at 

 each of the remaining throws ; and that of missing 

 an ace at the '.first throw, and throwing one ace, and 

 no more, in the two remaining throws. 



To estimate these, we must consider that the pro- 



bability of throwing an ace at one throw is - ; and 

 that of missing an ace in two successive throws, is 

 - ~~\ ' (Art. 17.); therefore the probability of 



throwing an ace at the first throw, and missing it at 



1 (n I) 1 . 

 the secpnd ana third throws, is - . N , (Art* 



Again, the probability of missing an ace at one 

 throw is 5 and that of throwing one ace, and 



2(n 1) . 



no more, m two successive throws, is --^ -' (Art. 



it* 



20-); therefore, the probability of missing an ace 

 the first throw, and throwing an ace, and no more, m 

 the two remaining throws, 15 



?fcii ( Art M ) -ifeir 



n* ^ '' w 3 



Taking now the sum of these probabilities, we have 



