738 



CHANCES. 



ghances. thereof may be expressed by the fraction f. Last'y, 

 s "*"Y"* il ^ he may reasonably expect \ of L. 5 as a consideration 

 for transferring his chance to another. 



PROBLEM III. 



10. Suppose that there are five counters, of which 

 four are black and one white,out of which, being mixed 

 together, a person blindfolded is to draw one, and is 

 to be entitled to L.5, or any other sum of money, if 

 he happens to draw the white counter. It is requi- 

 red to determine, before the event, what chance he 

 has of winning and of rrtissing the said L.5 ; and what 

 sum he may reasonably expect, upon transferring his 

 chance to another. 



SOLUTION. By reasoning in this as in the two 

 foregoing problems, it will appear, that the person 

 has only one chance in five of obtaining the L. 5, and 

 four chances out of five for missing it ; therefore, the 

 probability of his obtaining it may be expressed by 

 the fraction j- ; and the probability of his missing it 

 by the fraction . And he will be entitled to f of 

 the L. 5, if he transfers his chance to another. 



PROBLEM IV. 



11. Suppose there are five counters, three black, 

 and the other two white ; out of which, when mixed 

 together, a person blindfolded is to draw one, and is 

 to be entitled to five pounds, or any other sum, if he 

 happens to draw either of the white ones. It is re- 

 qufred to determine, before the event, what chance or 

 probability he 'has of winning, and missing, the said 

 five pounds ; and what sum he may reasonably expect 

 to be paid to him for resigning his chance to another. 



SOLUTION.- In this case, the person has manifestly 

 only two chances out of five for taking a white coun- 

 ter, and three for taking a black one : Therefore the 

 probability of winning may be expressed by the frac- 

 tion |., and that of missing by the fraction : Also 

 he ought to receive f of the five pounds if he parts 

 with his chance to another. 



12. What has been said in these four problems 

 concerning cards, counters, &c. may easily be con- 

 ceived to extend to any other things which are the 

 objects of chance. For instance, if, at the conclu- 

 sion of the drawing of a lottery, there should remain 

 in one wheel five tickets or numbers only, and in the 

 other, two equal prizes and three blanks ; then, the 

 possessor of one of these tickets would be exactly in 

 the situation of the person mentioned in the last ques- 

 tion. 



In general, if the number of blanks in a lottery be 

 represented by m, and the number of prizes by n, 

 then the probability of having one prize with one 



ticket will be - ; and the probability of having 



a blank 



m 



From which it appears, 



I. That the probability of the happening of any 

 vent, considered as resulting from chance, may be 

 expressed by a fraction, whose numerator is the num 

 ber of chances for the happening of the event, and the 

 denominator is the number of all the chances whereby 

 it may both happen and fail. And the probability of 

 such an event's failing, may be expressed by a frac- 

 jiqn whose numerator if the number of chances for its 



not happening, and denominator the same as that of f hanfter. 

 the former fraction. w-y-'w' 



II. The sum of the two fractions representing the 

 probability of the happening and the failing oj an 

 event is equal to unity : Therefore, when one of them 

 is given, the other may be found by subtraction. 



III. The expectation, that is, the sum which the 

 person who has a chance for the happening of an event 

 is entitled to, if he resign his chance to another) i* al- 

 ways the product of the fraction representing ike pro 

 bability multiplied into the sum expected: Therefore, 

 if that sum be denoted by unity, the expectation will 

 be denoted by the probability itself : Or, in general, 

 if the expectation of an event be divided by the value 

 of the thing expected, the quotient will express the 

 probability of the event. 



PROBLEM V. 



13. What is the probability that a person playing 3 

 with a single die throws an ace each time for two 

 successive throws ? 



SOLUTION. -Suppose that the person is to receive 

 L.I, provided he succeeds in throwing an ace each 

 time. Now, if an ace were to come up the first time, 

 then, because the die has six faces, only one of which 

 is favourable to him, his expectation on the second 

 throw would be xL.l, (Art. 12.) But we may 

 consider the first event, viz. the throwing an ace the 

 first throw as the condition of obtaining this sum 

 x L. 1 ; and the probability of this event being also 

 , the expectation before the first throw must neces- 

 sarily be X 7 X L. 1, (Art. 12.) ; and as in this case 

 the probability is expressed by the same fraction as 

 the expectation, the probability of throwing an ace 

 each time for two successive throws is |-x|-=-jV 



Corollary. As it is evidently the same whether a 

 single die be thrown twice successively, or two dice 

 be thrown at once ; therefore, the probability of 

 throwing two aces (or any given face) at one throw 

 with two dice, is also expressed by the fractiorf 



_ 



TJ* 



PROBLEM VI. 



14. A person offers to lay a wager of L.I, that 

 out of a purse containing m-{-n counters, whereof m 

 are black, and n white, he will, blindfolded, at the 

 first trial draw a white counter ; and also, that out 

 of another purse containing M + N counters, where- 

 of M are black, and N white, he will also at the first 

 trial draw a White counter, and that if he fails in ei- 

 ther trial his wager shall be lost. It is required to 

 determine the probability he has of succeeding there- 

 in. 



SOLUTION. If, as in the last problem, we sup- 

 pose he has already succeeded in the first trial, 

 then it will follow, as has been there explained, that 



N 

 his expectation on the second will be J^T^J X L. 1, 



N 



or simply rr -r;. And if, as in the same problem, 

 1 M-f-N 



we consider the success of his first trial as the con- 

 dition of obtaining this expectation, then its proba- 

 bility will be - ; which multiplied into that x- 

 * m-\-n 



