d'alembert. 291 



his chance of being right in the second case and wrong in all the 



others is 



7154321,,,. 1 



^X;iX^x-^XjX-Xx, that IS 



8 76 5 432' 8x6' 



and so on. 



If the meaning be that Paul undertakes to be right once at 



7 

 least in the seven cases, then his chance is - . For his chance of 



o 



being wrong every time is 



76543211 

 8^7^6^5^4^3^2' ^^ 8 ' 



therefore his chance of being right once at least is 1 — - , that is ^ . 



o 8 



Tlie second question means, I suppose, that Paul undertakes 

 to be right in the first two cases, and wrong in the other five. 

 His chance then is 



1154321,,,. ■» 



^XsX-X-XtX^Xt;, that IS 



8 7 6 5 4 3 2' 8x7x6* 



Or it may mean that Paul undertakes to be right in the first 

 two cases, but undertakes nothing for the other cases. Then his 



. 1 1 



chance is ^ x =■ . 



The third question means, I suppose, that Paul undertakes to 

 be right in two out of the seven cases and wrong in the other five 

 cases. The chance then will be the sum of 21 terms, as 21 combi- 

 nations of pairs of things can be made from 7 things. The chance 

 that he is right in the first two cases and wrong in all the others is 



1154321,. 1 



gX^x^x^x-x^x^, that IS 3 ^ ^ ^ g ; 



similarly we may find the chance that he is right in any two 

 assigned cases and wrong in all the others. The total chance will 

 be found to be 



8{7(6 + 5- + 4 + 3 + 2 + V + 6(5+4 + S + 2 + V 



19—2 



