Page:Supplement to the fourth, fifth, and sixth editions of the Encyclopaedia Britannica - with preliminary dissertations on the history of the sciences - illustrated by engravings (IA gri 33125011196181).pdf/653



So that the amount in this case, would be L. 4524, 1s, 7¼d.

II. ON THE PROBABILITIES OF LIFE.

134. Any persons A, B, C, &c. being proposed, let the numbers which tables of mortality (32) adapted to them, represent to attain to their respective ages, be denoted by the symbols $$a$$, $$b$$, $$c$$, &c.; while lives $$n$$ years older than those respectively, are denoted thus: $${}^n\!A$$, $${}^n\!B$$, $${}^n\!C$$, &c. and the numbers that attain to their ages, by the symbols $${}^n\!a$$, $${}^n\!b$$, $${}^n\!c$$, &c.; also let lives $$n$$ years younger than A, B, C, &c. he denoted thus: $$A_n$$, $$B_n$$, $$C_n$$, &c., while the numbers which the tables show to attain to those younger ages, are designated by the symbols $$a_n$$, $$b_n$$, $$c_n$$, &c.

Then, if A be 21 years of age, and we use the Carlisle Table, we shall have $$a = 6047$$, and $${}^{14}\!a = 5362$$, the number that attain to the age of thirty-five, or that live to be fourteen years older than A.

Hence the number that are represented by the table to die in $$n$$ years from the age of A, will be $$a-{}^n\!a$$, that is in 14 years, $$a-{}^{14}\!a$$; and by the Carlisle Table, in 14 years from the age of 21, that is, between 21 and 35, it will be 6047 − 5362 = 685.

135. Problem. To determine the probability that a proposed life A, will survive $$n$$ years.

Solution. $$a$$ being the number of lives in the table of mortality, that attain to the age of that which is proposed, conceive that number of lives to be so selected, (of which A must be one,) that they may each have the same prospect with regard to longevity, as the proposed life and those in the table, or the average of those from which it was constructed; then will the number of them that survive the term be $${}^n\!a$$ (134).

So that the number of ways all equally probable, or the number of equal chances for the happening of the event in question is $${}^n\!a$$; and the whole number for its either happening or failing is $$a$$; therefore, according to the first principles of the doctrine of probabilities, the probability of the event happening, that is, of A surviving the term, is $$\frac{{}^n\!a}{a}$$.

If the age of A be 14, the probability of that life surviving 7 years, or the age of 21, will, according to the Carlisle Table, be $$\frac{{}^7\!a}{a} = \frac{6047}{6335}$$, or 0·95454.

136. Since the number that die in $$n$$ years from the age of A is $$a-{}^n\!a$$ (134), it appears, in the same manner, that the probability of that life failing in $$n$$ years will be $$\frac{a-{}^n\!a}{a} = 1 - \frac{{}^n\!a}{a}$$ which probability, when the life, term, and table of mortality, are the same as in the last No. will be 0·04546.

137. If two lives A and B be proposed, since the probability of A surviving $$n$$ years will be $$\frac{{}^n\!a}{a}$$, and that of B surviving the same term will be $$\frac{{}^n\!b}{b}$$; it appears from the doctrine of probabilities that $$\frac{{}^n\!a}{a} \cdot \frac{{}^n\!b}{b}$$ or $$\frac{{}^n\!(ab)}{ab}$$ will be the measure of the probability that these lives will both survive $$n$$ years.

In the same manner it may be shown, that the probability of the three lives A, B, and C all surviving $$n$$ years, will be measured by $$\frac{{}^n\!a}{a} \cdot \frac{{}^n\!b}{b} \cdot \frac{{}^n\!c}{c}$$ or $$\frac{{}^n\!(abc)}{abc}$$. And, universally, that any number of lives A, B, C, &c. will jointly survive $$n$$ years, the probability is $$\frac{{}^n\!(abc, \&\text{c.})}{abc, \&\text{c.}}$$

138. Let $$\frac{{}^n\!a}{a} = {}_n\!a$$, $$\frac{{}^n\!b}{b} = {}_n\!b$$, $$\frac{{}^n\!c}{c} = {}_n\!c$$; also let $$\frac{{}^n\!(abc, \&\text{c.})}{abc, \&\text{c.}} = {}_n\!(abc, \&\text{c.})$$; so that the probabilities of A, B, C, &c. surviving $$n$$ years may be denoted by $${}_n\!a$$, $${}_n\!b$$, $${}_n\!c$$, &c. respectively; and that of all those lives jointly surviving that term, by $${}_n\!(abc, \&\text{c.})$$

Then will the probability that none of those lives will survive $$n$$ years, be $$(1-{}_n\!a)\,.\,(1-{}_n\!b)\,.\,(1-{}_n\!c)\,.\,\&\text{c.}$$

139. But the probability that some one or more of these lives will survive $$n$$ years, will be just what the probability last mentioned is deficient of certainty; its measure therefore, being just what the measure of that probability is deficient of unity, will be

140. ''Corol. 1. When there is only one life A'', this will be $${}_n\!{a}$$.

141. ''Corol. 2. When there are two lives A and B'', it becomes $${}_n\!{a} + {}_n\!{b}- {}_n\!(ab)$$.

142. ''Corol. 3. When there are three lives A, B'', and C, it becomes $${}_n\!a + {}_n\!b + {}_n\!c - {}_n\!(ab) - {}_n\!(ac) - {}_n\!(bc) + {}_n\!(abc)$$.

143. When three lives A, B, and C are proposed, that at the expiration of $$n$$ years there will be