(3.) Er. 3. If n balls a, b, c, d, &c. be thrown promiscuously into a bag, and a person draw out one of them, the probability that it 1 2 will be a is ; the probability that it will either be a or bis ?. n n 2 (6.) Er. 3. The same supposition being made, if two balls be drawn out, the probability that these will be a and b is 7.(n-1) n-1 Por there are n. combinations of n things taken two and two' together; and each of these is equally likely to be taken ; therefore 2 the probability that a and I will be taken is 7. (n-1): 1 a or n. 2 (7.) Er. 3. If 6 white and 5 black balls be thrown promiscuously into a bag, and a person draw out one of them, the probability that 6 this will be a white ball is and the probability that it will be a 11 a (8.) From the bills of Mortality in different places, tables have been constructed which shew how many persons, upon an average, out of a certain number born, are left at the end of each year, to the extremity of life. From such tables, the probability of the continuance of a life, of any proposed age, is known. (9.) Er. 1. To find the probability that an individual of a given age will live one year. Let A be the number, in the tables of the given age, B the number B left at the end of the year ; then is the probability that the indivi А A-B dual will live one year; and the probability that he will die А in that time. In Dr. Halley's Tables, out of 586 of the age of 22, 579 arrive at the age of 23 ? hence, the probability that an indi 579 83 7 1 vidual aged 22 will live one year is nearly; and 586 84 586' 84 nearly, is the probability that he will die in that time. (10.) Er. 2. To find the probability that an individual of a given age will live any number of years. Let A be the number, in the tables, of the given age, B, C, D,..X, B the number left at the end of 1, 2, 3,....t, years; then A с the probability that the individual will live one year; the probabi A or or IS X lity that be will live two years ; and the probability that he will А А m A-B A-C A-X live years. Also, are the probabilities that А he will die in 1, 2, t, years. These conclusions follow immediately from Art. 154. (11.) If two events be independent of each other, and the probability that one will happen be , and the probability that the other will happen, the probability that they will both happen is For each of the m ways in which the first can happen or fail, may be combined with each of the n ways in which the other can happen or fail, and thus form mn combinations, and there is only one in which both can happen ; therefore the probability that this will be the case 1 is (Art. 1). ( 1 . n mn (12). Cor. 1. The probability that both do not happen is I mn mn-1 or mn For, the probability that they both liappen, together with the probability that they do not both happen, is certainty ; therefore, if from unity, the probability that they both happen be subtracted, the remainder is the probability that they do not both happen. (13.) Cor. 2. The probability that they will both fail is (m-1)(n-1) For, the probability that the first will fail is m-1 . mn m n and the probability that the second will fail is therefore the n (m-1) (.n-1) probability that they will both fail is (14.) Cor. 3. The probability that one will happen and the other fail is m+1-2 For the probability that the first will happen and m+n-2, is the probability that one will happen and the other fail. (15.) Cor. 4. If there be any number of independent events, and 1 the probability that they will all happen is For, the proba mnt &c.' 1 bility that the first two will happen is and the probability that mn 5, and the chance of throwing a а 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. The chance of throwing an ace is 1 deuce in the second trial is therefore the chance of both hap 6 1 pening is 36. , 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 10' 1 because there are 5 white and 5 black balls left; therefore the 6 1 probability required is TiX 2 Or we may reason thus; unless 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 5 1 is To therefore the probability that both these events will take 6 3 place is Х or 2' 11 (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 3 (Art. 17.); when these two are removed, there are 5 11 white and 4 black balls left; and the probability of drawing a black or 3 or one is 4 3 vall, out of these, io ; therefore the probability required is ux g , is X 4 33 (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 and the chance of failing Õ 5 25 6' 36' 25 11 and the chance of not failing, both times, is 1 30 (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 the next is 5, therefore the chance of failing twice together is or 36* 5 of failing = times together is (3)", and this by the question is equal to the chance of happening, or (3)=1; hence, 1 x log. log.cz X X (log. 5. – log. 6) = log. 1 - log. 2; and x = Jog. 1 - log. 2 log. 2 since log. I=0; i. e. s=3.8, nearly. (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 , then the probability that they will both be alive at the end of that time is 1 be ; and the probability that Q will live a year, m n 1 X m 1 mn (22.) Et. 7. To find the probability that one of them, at least, will be alive at the end of any number of years. m-1 m n1 The probability that P will die in a year is and the probability that Q will die is *=*; therefore the probability that (m-1).(n-1) they will both die is and the probability that they m)n will not both die is 1 m+n-! (n-1).(n-1), or in n mn years, and , or a In the same manner, if be the probability that P will live P ? 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 (p-1).(9-1) end of the time is 1 . p+2= P 9 pa (23.) If the probability of an event's happening in one trial be represented by (Art. 3), to find the probability of its happening once, twice, three times, &c. exactly, in n trials. The probability of its happening in any one particular trial being ' the probability of its failing in all the other n-1 trials is lin (Arts. 3, 15.) therefore the probability of its happen(a+b)*ing in one particular trial, and failing in the rest is and (a+b)* since there are n trials, the probability that it will happen in soine one nalinof 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/24 n-1 all the rest, is and there are n(a + b)m' ways in which it may 2 happen twice in n trials and fail in all the rest ;(Com. Th .1.) therefore a ln-1 n. n. 2 the probability that it will happen twice in a trials is In (a+b) the same manner the probability of its happening exactly three times is n-1.1-2 2 (a + b)m ; and the probability of its happening exactly t times is 7-17-2 2 3 (a+b)* 3 natt Lalnas 7. (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 nh1.n-2 nt+1 an-LV 2 3 (a+b). n. |