36 the two first and third will happen is and the same proof may be extended to any number of events. When m=n=r&c. the probability is v being the number of events. (16.) Er. 1. Required the probability of throwing an ace and then a deuce with one die. 1 The chance of throwing an ace is k, and the chance of throwing a 1 deuce in the second trial is therefore the chance of both hap 5 1 pening is 17.) Er. 2. If 6 white and 5 black balls be thrown promiscuously into a bag, what is the probability that a person will draw out first a white, and then a black ball? 6 The probability of drawing a white ball first is (Art. 7), 11 5 and this being done, the probability of drawing a black ball, is vo' 1 6 3 X Or we may reason thus; unless 11 the person draw a white ball first, the whole is at an end; therefore the probability that he will have a chance of drawing a black ball 6 is and when he has this chance, the probability of its succeeding 11 ' 1 is therefore the probability that both these events will take or 5 or To 6 3 place is Х or 2' IT: (18.) Er. 3. The same supposition being made, what is the chance of drawing a white ball and then two black balls ? The probability of drawing a white ball and then a black one is (Art. 17.); when these two are removed, there are 5 11 white and 4 black balls left ; and the probability of drawing a black 3 3 mall, out of these, io q; therefore the probability required is X 4 4 933 (19.) Er. 4. Required the probability of throwing an ace, with a single die, in two trials. 5 The chance of failing the first time is Õ 25 the next is 36' 25 11 and the chance of not failing, both times, is 1 30' 36° (20.) Er. 5. In how many trials may a person undertake, for an even wager, to throw an ace with a single die? Let be the number of trials; then, as in the last Art. the chance is 5, and the chance of failing 6, therefore the chance of failing twice together is or log or x x (lug. 5. – log. 6) = log. 1 - log. 2; and r = Ing. 1 - log. 2 log. 2 since log. I=0; i. e. r=3.8, nearly. log. 5-log. 6 log. 6-log. 5 (21.). Er. 6. To find the probability that two individuals, P and Q, whose ages are known, will live a year. Let the probability that P will live a year, determined by Art. 9, 1 be 1 1 Х m R OT probability that they will both be alive at the end of that time is 1 mn (22.) Er. 7. To find the probability that one of them, at least, will be alive at the end of any number of years. m-1 The probability that P will die in a year is and tho m In the same manner, if be the probability that P will live P 1 1 years, and the probability that Q will live the same time (Art. 9 10); the probability that one of them, at least, will be alive at the end of the time is 1 (p-1).(9-1), or P+q Pa (23.) If the probability of an event's happening in one trial be represented by (Art. 3), to find the probability of its hap a to pening once, twice, three times, &c. exactly, in n trials. The probability of its happening in any one particular trial being atr' the probability of its failing in all the other n-1 trials is a a nalyn (Arts. 3, 15.) therefore the probability of its happen(a+b)-ing in one particular trial, and failing in the rest is a ln-1 ; and (a+b)* since there are n trials, the probability that it will happen in soine one of these, and fail in the rest, is n times as great, or The (a+b) probability of its happening in any two particular trials and failing in a/3all the rest, is n-1 ways in which it may 2 happen twice in n trials and fail in all the rest ; (Com. Th.1.) therefore n-1 n. Laolo 2 the probability that it will happen twice in n trials is In (a+b) the same manner the probability of its happening exactly three times is n-1.1--2 2 3 (a + 6)" ; and the probability of its happening exactly t times is (a + b)wo and there are n n. n-1.1-2 not+1 2 3 t (a+b) (24.) Cor. 1. The probability of the event's failing exactly t times in n trials may be shewn in the same way, to be n-1.1-2 net+1 -an2 3 (a+b)* n. 2 23.) Cor. 2. The probability of the event's happening at least 1 imes in n trials is N-1 a" + nan-16+ n. an=268 .. to n-+1 terms 2 (a+b). For if it happen every time, or fail only once, twice,....no it happens t times; therefore the whole probability of its happening at least t times, is the sum of the probabilities of its happening every time, of failing only once, twice,....-t times ; and the sum of these probabilities is n-I a*+nan-'1+n. an-262..., to nt+terms. (a+b)* (26.) Er. 1. What is the probability of throwing an ace, twice, at least, in three trials, with a single die? In this case, n=3,1=2, a=), b=5; and the probability required 1+3.5 16 2 is 6.6.6. 216 27 (27.) Er. 2. What is the probability that out of 5 individuals, of a given age, three, at least, will die in a given time. 1 Let be the probability that any one of them will die in the given time (Art. 10).; then we have given the probability of an event's happening in one instance, to find the probability of its happening three times in five instances. In this case, a=1, l=m-1,n=5, t=3, therefore the probability 1+54.m-1)+10.(m-1) required is (27.) Scholium. Much more might be said on a subject so extensive as the doctrine of chances; the Learner will, however, find the principal grounds of calculation in the 1, 3, 11, 23, and 29 Articles, and if he wish for farther information, he may consult De Moivre's work on this subject. It may not be improper to caution him against applying principles which, on the first view, may appear self-evident, as there is no subject in which he will be so likely to mistake as in the calculation of probabilities. A single instance will shew the danger of forming a hasty judgment, even in the most simple case. Tho probability of throwing an ace with one die is a, and since there is an equal probability of throwing an ace in the second trial, it might be supposed that the probability of throwing an ace in two trials is This is not a just conclusion (Art. 19); for, it would follow, by the same mode of reasoning, that in six trials a person could not fan to throw an ace. The error, which is not easily seen, arises from lacit supposition that there must necessarily be a second trial, which is not the case if an ace be thrown in the first. m ON LIFE ANNUITIES. (1.) To find tne present value of an annuity of €1, to be continued during the life of an individual of a given age, allowing compound interest for the money. Letr be the amount of £i, in one year ; A the number of persons, in the tables, of the given age; B, C, D, &c. the B aumber left at the end of 1, 2, 3, &c. years' ; then is the value A C D of the life for one year, &c. its value for 2, 3, &c. years; A' A and the series must be continued to the end of the tables. Now the present value of £1, to be paid at the end of one year is ; but it is only to be paid on condition that the annuitant is alive at the end of B the year, of which event the probability is therefore the preA! B sent value of the conditional annuity is in the same A с ner, the present value of the second year's annuity is the Ari pre D sent value of the third year's annuity is Anni &c. therefore the ; man ; whole value required is À BX+om+&c.) to the end of the tables. (2.) De Moivre supposes, that out of eighty-six persons born, one dies every year, till they are all extinct. This supposition is sufficiently exact, if our calculations be made for any age above ten, as will appear from an inspection of the tables ; and on this supposition, the sum of the series B с D + Let n be the number of years which any individual wants of 80 ; |