ESSAY ON PROBABILITIES. 



BY TABLE I. 



Divide p by p + g, and call- 

 ing the result II, find the cor- 

 responding value of t in the 

 table. Multiply it by the 

 square root, subtract 1 and 

 divide by 2 ; the quotient 

 being called Z, it is then p to q 

 that the As in n (a + b) trials 

 shall be contained between 

 nal and na + l, both inclu- 

 sive. 



BY TABLE II. 



Find the square root ; mul- 

 tiply by 31, and divide by 

 130. Divide p by p+q, and 

 calling the quotient K, find the 

 corresponding value of t in 

 the table ; multiply t by the 

 pieceding quotient, subtract '5 

 from the product, and I being 

 the remainder, it is then p to q 

 that the As in n (a-t-fc) trials 

 shall be contained between 

 nal and na + 1, both inclusive. 



I shall conclude this problem with an example of each 

 case, worked by both methods, without explanation. 

 There is a lottery containing 3 white and 2 black balls : 

 what is the chance that in 50,000 drawings the num- 

 ber of white balls shall be between 30,000 + 100 and 

 30,000-100? 



fl-3 6 = 2, n=10,000, 7=100 

 8 x 3 x 2 x 10,000 



= 96,000 



V 96,000 =309 -84 



309 -84) 201 ( -64873 = t 

 H= -64109 



309-84 

 31 



130) 9605 -04(73 -885 



73-885)100'5(l-3602 

 K= -64109 



This question shows how nearly a great many trials 

 may be expected to agree with the probable mean : in 

 50,000 trials, it is nearly two to one against the 

 number of white balls differing from 30,000 by more 

 than a hundred. 



In 1 00,000 tosses, between what limits is it 99 to 1 

 that the heads shall be contained ? 



l, 6=1, n = 50,000,^ = 99, j=l 

 100)99('99 = H 

 t^l-8215 



8 x 1 x 1 x 50,000 



-=200,000 



447-21 

 31 



130)13863-51(106-64 

 100)99(-99-=K 



