Hence, since there is only one chance for his succeeding, while there are four for his failing, the odds against him will be as four to one.
Prob. 2. To determine the probability that an event happens a given number of times in a given number of trials, sup posing, as in the former problem, the probability of its happening each time to that of its failing to be in the ratio of a to b.
Sol.' It will be observed that this pro blem materially differs from the preced ing, in as much as the event in that pro blem was restrained, so that it should happen neither more or less often than a given number of times, while in this pro blem the event is determined equally fa vourable by its happening either as often or oftener than a given number of times, so that in the present case there is no further restriction than that it should not fall short of that number.
1. Let the probability be required of an event happening once at least in two trials. If it happens the first and fails-the second time, or fails the first and happens the second time, or happens both times, the event will have equally succeeded. The probability in the first case is ;, the l b a in the second is n the a probabilityinthethirdis-L--a ' ; hence the (1-1-61 2ab-l-aa probability required will be 2. Let the probability be required of its happening once in three times. Pro vided it has happened once at least in the first two trials, the event will have equally succeeded, whether it happens or fails in a'±2ab the third trial, and therefore will represent the probability in this case. But it may have failed in the first two and hap pened in the third trial, the probability of b b a which ; adding this to the preced ingfraction we b X a+ a a3-1-3eb-1-3ab' for the probabilitya -OP required. In like manner the proba bility of its happening once at least in four trials will be a3 ± 3 a b 3a b b a + a b3 a3 a' P-1-4 a b3, and FA 4 the probability of its happening once at n least in n times will be = a+bn b -. In a1-7)" other words, since the event must happen once at least, unless it fails every time, the probabilityrequired (by D ef. 1)will always be expressed by the difference between be unity and .
a-M-b 3. Let the probability be required of an event's happening twice at least in three trials. In this case it will succeed, if it hap pens the first and second, and fails the third time, if it happens the first and third, and fails the second time, if it happens the second and third, and fails the first time, or if it happens each time successively. The firstthree probabilities and 03 the fourth is therefore the proba a + bl 3 bility required will be a3 ab If the event is to happen twice at least in four times, the probability of its happening ing the first three times has been already found. Let it be supposed to have ed only once in three times, the probability of which, by the preceding problem, in 3ab_b then will the probability of its hap pening the fourth, after having happened a' bl in the three preceding, be a -1-b) and therefore the whole probability will be a3-1-3 a' b, 3 a' a4-I-4a3 a + b a 4 By proceeding in the same manner, it may be found that the probability of an event's happening twice at least in five trials, will be a3 b+6 a' a 4 a b3 X - a b a+b Tp, 4aS-F5a4 a3 /04-10 03 61s And if the probability of the event's happening thrice in 4, 5, 6, &c. trials be required, they may, by pursuing the same steps, be 4 a3 b. as-}- 5 b 10 a3 a a b 6+15 a3 63 &c.
Fb 1 re spectively. Hence it follows, that if the binomial a be raised to nth power, the probability of an event's happening at least d times in n trials will be = a b2 1a) a -Hflb that is, the series in the numerator must be continued till the index of a becomes to d.
Cor. From this solution it appears that the series.
n-1 bn to d terms, will express the probability of the event's not happening so often as d times in n trials.
Ex. Supposing a person with six dice undertakes to throw two aces or more in the first trial, what is the probability of his succeeding? In this case a, 6, n, and d, being respectively equal to 1, 5, 6, and 2, the above expression will be come = 1+30±15 X25-1-20 x 125+15><625 Hence the odds against his succeeding will be as 34375 to 12281, or nearly as three to one.