Prob. 1. To determine the probability that an event happens a given number of times, and no more, in a given number of' trials.
Sol. 1. Let the probability he required of its happening only once in two trials, and let the ratio of its happening to that of its failing be as a to b. Then, since the event can take place only by it happen ing the first, and failing the second time, the probability of which X a b a + b = or by its failing the first and happening the second time, the probability of which , the sum of a a b these two fractions, or2 s will be a bl the probability required.
2. Let the probability be required of its happening only twice in three trials. In this case, the event, if it happens, must take place in either of three different ways: 1st, by its happening the first two, and failing the third time, the probabili a a b ty of which is 2dly, by its failing the first, and happening the other two times, the probability of which is b a a ,; or, idly, by its happening the a + b ' first and third, and failing the second a time, the probability of which is a + . b , a The sum of these fractions, therefore, or will be the required probabili a b ty. By the same method of reasoning, the probability of its happening only once in three trials, or, which is the same thing, of its failing twice in three 3 b b trials, may be found equal to a 3. Let the probability of the event's happening only once in four trials be re quired. In this case it must either hap pen the first and fail in the three suc ceeding trials ; or happen the second and fail in the first, third, and fourth trials ; or happen the third, and fail in the first, second, and fourth trials ; or happen the fourth, and fail in the first, second, and third trials. The probability of each of these being 4, the required proba • 4ab3 bility will be 4; and for the same rea a±b • son, the probability of its happening three times and failing only once in four trials 4 b a3 will be a + b) 4. Let the probability be required of its happening twice and failing twice in four trials : here the event may be deter mined in either of six different ways 1st, by its happening the first and second, and failing in the third and fourth trials; 2dly, by its happening the first and third, and fidling the second and fourth trials; idly, by its happening the first and fourth, and failing the second and third trials ; 4thly, by its happening the second and third, and failing the first and fourth trials ; 5thly, by its happening the second and fourth, and failing the first and third trials ; or, 6thly, by its happening the third and fourth, and failing the first and second trials. Each of these probabili
ties being expressed by follows a -I- bl that the sum of them, or 4 will ex a±bl press the probability required.
By proceeding in the same manner, the probability in any other case may be determined. But if the number of trials be very great, these operations will be come exceedingly complicated, and there fore recourse must be had to a more ge neral method of solution.
Supposing n to be the whole number of trials, and d the number of times in which the event is to take place, the probability of the event's happening d times succes sively, and failing the remainingn d times, will be x • a + 6) b a ± b) But as there is the same probability of its happening any other d assigned trials and failing in the rest, it is evident that this probability ought to be repeated as often as d things can be combined in n things, which, by the known rules of combina it n-1 n-2 . , tion, are = x X continued 1 2 3 to d terms; the general rule therefore ad will bea multiplied into n X ± n-1 n-2 n-3 2 X .„ a X— 4 continued to .d terms.
Er. Supposing a person with six dice undertakes to throw two aces, and no more ; or, which is the same thing, that he undertakes with one die to throw an ace twice, and no more, in six trials ; it is required to determine the probability of his succeeding, a being in this case = 1, b= 5, n = 6, and d = 2, the above expres 54 sions will become —, multiplied into 5 4 625 66 151 5 6 X - 2 = = very nearly.