probability it rests upon Fb added to the probability it dont, as fb to A B, or as the ratio of fb to Ab to the ratio of AB to AB. But the probability of any event added to the probability of its failure is the ratio of equality; wherefore he probability if rest upon Fb is to the ratio of equality as the ratio of fb to AB to the ratio of AB to AB, or the ratio of equality; and therefore the probability it rest upon Fb is the ratio of fb to AB. But er hypothesi according as the ball W falls upon Fb or nor the point o will lie between f and b or not, and therefore the probability the point o will lie between f and b is the ratio of fb to AB Again; if the rectangles Cf, Fb, LA are not commensurable, yet the last mentioned probability can be neither greater nor less than the ratio of fb to AB: for, if it be less, let it be the ratio of fc to AB, and upon the line fb take the points p and t, so that pt shall be greater than half cb, and taking p and t the nearest points of division to f and c that lie upon fb). Then because Bp, pt, tA are commensurable, so are the rectangles Cp, Dt, and that upon pt compleating the square AB. Wherefore, by what has been said, the probability that the point o will lie between p and t is the ratio of pt to AB. But if it lies between p and t it must lie between f and b. Wherefore, the probability it should lie between f and b cannot be less than the ratio of fc to AB(since pt is greater than fc) And after the same manner you may prove that the forementioned probability cannot be greater than the ratio of fb to AB, it must therefore be the same Lem. 2. The ball W having been thrown, and the line os drawn, the proba bility of the event M in a single trial is the ratio of Ao to AB For, in the same manner as in the foregoing lemma, the probability that the ball o being thrown shall rest somewhere upon Do or between AD and so is the ratio of Ao to AB. But the resting of the ball o between AD and so after a single thrwo is the happening of the event M in a single trial. Wherefore the lemma PROP 8 If upon BA you erect the figure BghikmA whose property is this, that(the base ba being divided into any two parts, as Ab, and Bb and at the point of division b a perpendicular being erected and terminated by the figure in m; and y, I, r representing respectively the ratio of bm, Ab, and Bb to AB, and e being the coefficient of the term which occurs in aPbi when the binomial a+ bp+q is xpanded )y= ErPri. I say that before the ball W is thrown, the probability and withall that the event M should happen p times and fail q in p+q trial the point o should fall between f and b, any two o=points named in the line perpendiculars fg, bm raised upon the line AB, to Ca the square upon Ap e t the ratio of f ghikmb, the part of the figure Bghikm A intercepted between the DEMONSTRATION or if not; lst let it be the ratio of d a figure greater than fghikmb to CA, nd through the points e, d, c draw perpendiculars to fb meeting the curve AmigB in h, i, k; the point d being so placed that di shall be the longest of the perpendiculars terminated by the line fb, and the curve Amig B; and the pointsprobability it rests upon Fb added to the probability it dont, as f b to A B, or as the ratio of f b to AB to the ratio of AB to AB. But the probability of any event added to the probability of its failure is the ratio of equality; wherefore, the probability if rest upon Fb is to the ratio of equality as the ratio of f b to AB to the ratio of AB to AB, or the ratio of equality; and therefore the probability it rest upon Fb is the ratio of f b to AB. But ex hypothesi according as the ball W falls upon Fb or nor the point o will lie between f and b or not, and therefore the probability the point o will lie between f and b is the ratio of f b to AB. Again; if the rectangles Cf, Fb, LA are not commensurable, yet the last mentioned probability can be neither greater nor less than the ratio of f b to AB; for, if it be less, let it be the ratio of f c to AB, and upon the line f b take the points p and t, so that pt shall be greater than half cb, and taking p and t the nearest points of division to f and c that lie upon f b). Then because Bp, pt, tA are commensurable, so are the rectangles Cp, Dt, and that upon pt compleating the square AB. Wherefore, by what has been said, the probability that the point o will lie between p and t is the ratio of pt to AB. But if it lies between p and t it must lie between f and b. Wherefore, the probability it should lie between f and b cannot be less than the ratio of f c to AB (since pt is greater than f c). And after the same manner you may prove that the forementioned probability cannot be greater than the ratio of f b to AB, it must therefore be the same. Lem. 2. The ball W having been thrown, and the line os drawn, the proba￾bility of the event M in a single trial is the ratio of Ao to AB. For, in the same manner as in the foregoing lemma, the probability that the ball o being thrown shall rest somewhere upon Do or between AD and so is the ratio of Ao to AB. But the resting of the ball o between AD and so after a single thrwo is the happening of the event M in a single trial. Wherefore the lemma is manifest. P R O P. 8. If upon BA you erect the figure BghikmA whose property is this, that (the base BA being divided into any two parts, as Ab, and Bb and at the point of division b a perpendicular being erected and terminated by the figure in m; and y, x, r representing respectively the ratio of bm, Ab, and Bb to AB, and E being the coefficient of the term which occurs in apbq when the binomial a + b| p+q is expanded) y = Exprq. I say that before the ball W is thrown, the probability the point o should fall between f and b, any two o=points named in the line AB, and withall that the event M should happen p times and fail q in p + q trials, is the ratio of fghikmb, the part of the figure BghikmA intercepted between the perpendiculars fg, bm raised upon the line AB, to CA the square upon AB. D E M O N S T R A T I O N. For if not; 1st let it be the ratio of D a figure greater than fghikmb to CA, and through the points e, d, c draw perpendiculars to f b meeting the curve AmigB in h, i, k; the point d being so placed that di shall be the longest of the perpendiculars terminated by the line f b, and the curve AmigB; and the points 9