PROBLEM 1.-The probabilities of r and Q, at. any one trial, are as a to b, that is, it is a to b for against o, and b to a for q against P. A large number of trials n(a+ b) is made; what are the chances that the number of PS which happen shall lie between n and n a-1, and (of course) the number of Qs between st b-1 and nb+1 (1 being small compared with n a or n b)l RULE.-Calculate.-Calculate (21+1)-a- let this be the A of the a+ b table (altered as above); then the corresponding B, altered as above, is the probability required.
EXAMPLE.-It is 2 to 1 that a throw with a die shall give 3, 4, 5, or 6, and not 1 or 2 ; what is the chance that, in 12,000 throws, the number which gives 3, 4, 5, or 6 shall lie between 8000 +100 and 8000-100. Here a=2, b=1, n=4000, 1=100, and the complete calculation by logarithms • (which we insert merely as a guide to the readiest mode of treating more complicated cases) is as follows In the table, A being P38, is which is more than near enough for the purpose, my : the result then is that .95 is the chance in favour of the throws which give 3, 4, 5, or 6 out of 12,000 throws lying between 8000 + 100 and 8000 -100, whence .05 is the chance against it ; that is, it is about 05 to 5, or 19 to 1, that the throws shall eo lie. A few instances of this kind will show bow com pletely, in the long run, events may be expeeted to happen in numbers nearly proportional to what are called, when the precedent circum dances are fully known, their probabilities of happening.
Pnontzar 2.-1n the preceding problem, to find 1 so that there may be a given probability that the rs shall lie between n and n a-/.
Rum-Reduce the fraction which represents the given probability to a decimal fraction, and take the A corresponding to the is which is nearest to this fraction in the table ; multiply by the square root above described, subtract 1, and divide by 2; the whole number nearest to the result is the answer required for 1.
ExAstrtE.-In the preceding example, find 1 so that it is an even chance that the number of throws out of 12,000 which give 3, 4, 5, or 6 shall lie between 8000-i and Here f, or .5 is the given probability, the nearest to it in column B is to which A is .48, which multiplied by the square root (or the number whose logarithm is 2'1646) is subtract 1 and divide by 2, which gives 33 very nearly : hence it is an even chance (very nearly) that the number of throws out of 12,000, which give 3, 4, 5, or 6 lie between 8000-33 and 8000 + 33.