"ber of combinations would be 1 + 2 + 3; so with another letter, we would have 1+ 2+ 3+ 4 combinations; with m letters 1+ 2+ 3+ 4+ 5, &c.tom-1.The sum of this series is m letters; which is according to the rule above given. 3. If we have 3 letters a, b, c, and wish to combine 3 at a time, we can have but one combination, abc. Now if we have another letter d, we shall have 3 more different combinations of 3 letters; for, we can place d in place of each one of the letters abc. Hence the number of combinations, of 3 out of 4, will be 13. If we have another letter e, we shall have 6 more combinations; for, there are 6 combinations of 2 letters in the 4, a, b, c, d; and with each one e may be placed. Hence the number of combinations of 3 in 4, is 1+3+6. In a similar way, the number of combinations of 3 out of 5, may be shown to be 1+3+6+10 and the number of combinations of 3 out of 6, may be shown to be 1+3+6+10+15, &c. And we find, by the method of differences, the sum of this series to be m being the number of letters out of which the combination is made.. In the same way the number of combinations of 4 out of m, may be shown to be And, in general, the number of combinations of n out of 4. What number of products can be made of the 4 numbers, 3, 4, 5, 6, taking them 2 by 2? 5. What number of combinations can be made of 6 letters out of 24? 6. How many syllables of 3 letters can be formed out of 18 letters, no one of which occurs twice in the same syllable; and no two syllables containing all the same letters? 7. How many different combinations of 2 letters can be made by selecting out of the 24 letters of the alphabet? SECTION CIX. If a coin be thrown up, and it is required what will express the probability that one particular face will turn up; since there are but two faces, and one or the other must turn up; and, since there is no reason why one should turn up in preference to the other, the chance of either one turning up is, or the chance of one turning up is, and the chance of its not turning up is . Add together the two fractions expressing the chance of a particular face happening to turn up, and failing to turn up, and they make unity. A die has six faces. If a die be cast, what is the chance that a particular face will turn up? It is evident that the chance is only, since there are six faces, and there is no reason why any particular face will turn up. The chance against any particular face turning up is. The fractions which express the chance of the face happening to turn up, and failing to turn up, when added together, are equal to unity. 1. There are 4 white balls and 6 black ones; what is the chance of drawing out at hazard, a white ball or a black one? 2. If 600 dollars depended on the turning up of a particular face of a die; what would the chance be worth? 3. If m represent the number of chances on one side, and n the number on the opposite side; what will represent the chance on either side? m 4. What is the chance at a particular face of a die will turn up at 2 successive timyys? First, the chance that a particular face will turn up at the first throw is only ; hence, the chance of making the second trial, is only; and the chance of succeeding at the second trial, is only. We have then, only of of a chance of succeeding at the two successive throws, or of a chance of success. Upon the same principle the chance of any particular face turning up at 3 successive throws, is of, or st 5. If there be 4 white balls in a hat and 6 black ones; what is the chance of drawing out 2 black balls at two successive trials? First, the chance of drawing a black ball at first, is only 4 out of 10, or. The chance of making the second trial at all, is only ; but, the chance of succeeding at the second trial, is only 3 out of 9, or 3. The chance of success at both trials is then of 3, or 13. 6. What is the chance of drawing out the 4 black balls, at 4 successive trials? 7. What is the chance of drawing out 2 white balls at 2 successive trials? 8. What is the chance of drawing out the whole 4 at 4 successive trials? 9. What is the chance that every face of a die will turn up at 6 successive throws? SECTION CX. To determine the chance of an event happening a given number of times and no more, in a given number of trials, when there are m chances for its happening, and n for its failing. 1. It is required to ascertain the chance of an event happening once, and no more, in 2 trials. Supposing the chance for its happening be m, and the chance for failing n, each time; then the case proposed may happen two ways; it may happen the first trial, and fail the second, or it may fail the first, and happen the second; but, the chance of its happening the first time is W m m+n and of fail. failing the first time, and happening the second, is m n m + n therefore, the chance of the event happening once, and only 2. What is the chance for the event happening once, and only once, in 3 trials? 3. What is the chance for the event happening once in 4 trials? 4. What is the chance for the event happening twice in 3 trials? Three times in 5 trials? 5. What is the chance for throwing an ace twice, and no more, in 6 trials? SECTION CXL Chances. To find the probability of an event happening at least a given number of times, without limiting the number, the chance of its happening and failing each time being given. 1. It is required to find the chance of an event happening once in 2 trials, when the chances of its happening each time are m, and the chances of failing n. This may take place three ways.; it may happen the first and fail the second, or happen the second, and fail the first, or it may happen at both trials. Hence the chance is 2 m n + m2 (m + n)3 2. What is the chance for the event happening twice in 3 trials? Twice in 4 trials? Three times in 5 trials? Three times in 6 trials? SECTION CXII. Chances. 1. Suppose that 3000 children die out of 11650, in 1 year after birth; what is the chance that an infant will live 1 year? 2. One person is 50 years of age and another 60. Between the ages of 50 and 60, 82 persons die annually out of every 2857. Between the ages of 60 and 70, 80 die annually out of every 1712. What is the chance that either will live 10 years? What is the chance that the persons mentioned will both live 10 years? What is the chance that the first will live 10 years; and that the other will not? 3. A person 30 years of age has an annuity for 10 years, the present worth of which is 1000 pounds, provided he lives but the 10 years; for, if he dies, the annuity ceases. What is the annuity worth, as it is ascertained that about 75 out of every 4385 persons die annually, between the ages of 30 and 40 years? 4. A, who was 70 years of age, had an annuity which was to last 10 years, provided he lived to the end of that time. B gave him for it, at a fair price, 1250 dollars; but, he has forgotten what amount A was to recieve annually. Now, between the ages of 70 and 80, 80 persons die out of every 832 upon an average. What then was A's annuity worth, in hand, provided his life had been secured for 10 years? 5. A, B and C, who are respectively 30, 40 and 50 years of age, are to receive a sum of money at the end of 10 years, provided they live until that time, which they are to share equally; but, if A dies, B and C are to have, each, of A's in addition to his own. If B dies, A and C are to have, each, of his in addition to his own. If C dies, A and B are to have, each, of his in addition to his own; if any two of them die, the remaining one is to have of their shares. The whole of their chances is worth 1260 dollars, payable at the end of 10 years. What was the sum of money which each was to receive, supposing 75 die out of every 4010, between the ages of 30 and 40; 82 out of every 2448, between the ages of 40 and 50; and 80 out of every 1632, between the ages of 50 and 60? |