CHAPTER II THE THEORY OF PROBABILITY 14. Fundamental Principle.—Originating at the gaming table in the middle of the seventeenth century and developed by the brilliant school of French mathematicians that flourished for the next 150 years, the theory of probability has come to occupy a place of fundamental importance in very diverse fields of human inquiry. Today the physicist, the chemist, the statistician, the biologist, the actuary, depend in ever-increasing measure upon the results of this theory for the further development of their respec- tive sciences. The theory of probability is based upon the following funda- mental principle, or definition: If an went can happen in a ways and fail to happen in b ways, all being equally likely, the probability that it will happen in any one trial is p=———» (1) a + 6 and the probability that it mil fail to happen in any one trial is q = ——— (2) - a + b Whether the events are all equally likely or not can be determined, in general, only by experiment. For example, in throw- ing a cubical die, the presumption is that the six possible ways in which the die can fall are all equally likely. Since the ace can fall uppermost in one way and can fail to fall uppermost in five ways, the probability of getting ace in a single throw is one in six, and the probability of not getting ace is five in six. But if, in a large number of trials, the ace should fail to fall uppermost in approximately one-sixth the number of trials, the presumption would be that the six possible ways in which the die can fall are 13 14 MATHEMATICS OF LIFE INSURANCE [§15 not all equally likely and that the die is either imperfect or else the method of throwing is not a matter of pure chance. Many games of chance are like the example just quoted; that is, it is possible to determine a priori, or beforehand, in how many ways a given event can happen and in how many ways it can fail to happen. Exercises 1. Ten cards are numbered from 1 to 10 and one is drawn at ran- dom. What is the probability that it is No. 5? 2. Two dice are thrown together. What is the probability that both fall ace uppermost ? Of getting a count of 7 ? Of getting a count of 4 or a count of 7? 3. A bag contains three white balls, two black balls, and five red balls. One ball is drawn at random. What is the probability that it is red? Black? White? 4. In the preceding exercise, two balls are drawn. (a) How many groups of two in ten? Of two in three? Of two in five? (&) What is the probability that both balls are black? That both are white? That both are red? That one is white and one is red? 5. In Exercise 3, three balls are drawn at once. What is the prob- ability that all are red? 6. Four cards are drawn at random from a deck of fifty-two cards. What is the probability that they are the four aces? That there is one card from each suit? 15. Bernoulli's Theorem.—In many applications of the theory of probability it is not possible to determine a priori in how many ways a given event can happen. For example, if an um contains a large number of small balls and we require the probability of drawing a black ball in a single trial, having no means of determin- ing o priori how many balls there are in the urn, or even if there are any black balls in it, the problem can only be solved by experi- ment. The experiment is performed by the process of "sampling." A handful of balls is withdrawn and the result observed. Suppose it is found that there are twice as many white balls as black balls in the handful. We should then only know that there are white balls and black balls in the um. But if, after a great many such trials, obtaining each time only white §15] THE THEORY OF PROBABILITY 15 balls and black balls in approximately the ratio 2:1, the presump- tion would be that the urn contains only white balls and black balls in approximately the ratio 2:1, and that the probability of drawing a black ball in a single trial is approximately one in three. The application of the theory of probability to problems in natural science, statistics or actuarial science resembles the problem just considered. The fundamental principle of the preceding section is modified to suit such problems and is known as Bernoul- li's Theorem If a given event happens r times in n trials, n being very great, then the probability that the given event will happen in any one trial is P-^' W and the probability that the given event will fail to happen in any one trial is1 n — r ,,. q = -n- (2) As an example of the application of this modified principle, suppose it is known by observation that, out of 100,000 children, each ten years of age, 69,804 will be alive at age fifty. The probability that any one child of age ten will live to be fifty is 69,804 p = 100,000 = ()-69804 and the probability that any one child of age ten will fail to reach age fifty is '-S-»-30196. since the event of living has happened 69,804 times in 100,000 trials. A glance at the fundamental principle, or its modified form, shows at once that both p and q are fractions ranging in value from Oto 1. ' More accurately expressed, p ° ^T«^' ? = ^T»—^— A proof will be found in Sec. 19. 16 MATHEMATICS OF LIFE INSURANCE [§16 If p = 0, the event in question cannot happen, or is impossible. If p = 1, the event in question is certain to happen. Again, it follows at once that P + q = 1, (3) an equation that may be thought of as symbolizing the truism that an event must either happen or fail to happen. 16. Compound Probability.—The probability that a series of independent events will happen simultaneously or in sequence is called compound probability. The determination of compound probability, when the probability of the occurrence of each event taken by itself is known, rests upon the following theorem: Theorem.—The probability that a series of n independent events will happen simultaneously, or in sequence, is the product of the separate probabilities that each event will happen when taken by itself. If we denote by 0,1, a,, as, . . . , On the number of ways the respective events of the series can happen, and by 61, bi, bs, . . ., bn the number of ways the respective events of the series can fail to happen, then the total number of ways in which these events can happen simultaneously, or in sequence, is the continued prod- uct di 02 ay . . . . a, (cf. Sec. 10); and the total number of possibilities, favorable and unfavorable, is the continued product (ai + 6i)(ff2 + &2)(ff3 + bs) . . . (o» + b»). Hence, by defini- tion (Sec. 14), the required probability is fli" da' as . . . a,n _____ p = (Ol + 6l)(ffi2 + &2)(03 + 63) ... (On + &„) = Pi ?2 PS . . . Pn, (1) where pi, pi, ps, . . , pn indicate the respective probabilities that the events will happen when each is taken by itself. This theorem has a number of important consequences, as follows: Consequence 1.—The probability that a series of n independent events will all fail to happen is q = 3i <?2 qs 94 . . . qn = (1 - pi)(l - pa)(l - Pa) (1 - p4) ... (1 - P»), (2) where q\, qi, qs, q^, . . . , q» indicate the respective probabilities that the events will fail to happen when each is taken by itself. $16] THE THEORY OF PROBABILITY 17 Consequence 2.—The probability that not all the events of a series of n independent events will happen simultaneously, or in sequence, that is, that at least one will fail to happen, is 1 - p = 1 - pi pi ps p4 . . . p». (3) Consequence 3.—The probability that not all the events of a series of n independent events will fail to happen, that is, that at least one event mil happen, is 1 - q = 1 — qi qi qs 94 . . . <?„. (4) Consequence 4.—The probability that some particular group of r events out of a series of n independent events will happen and the remaining n — r events will fail to happen is nP, = PrQn-r, (5) where Prdenotes the compound probability that all the events of the particular group of r events will happen and Qn-r indicates the compound probability that all the remaining n — r events will fail to happen. For example, if the n independent events are arranged in any definite order, the probability that the first r events will happen and the remaining n — r events will fail to happen is Pi Pi ?3 Pr <?r+l 9r+2 ?r+3 . . . ffn. Problems involving compound probability are indicated by the word "and," either expressed or implied. Thus, if it is known that the probability a man will live to celebrate his golden wedding is three in four and the probability that his wife will live to celebrate the same anniversary is two in three, then the probability that both (implying the man and his wife) will live to celebrate this anniver- sary is M X % = %; the probability that neither will live (implying the man will fail to live and his wife will fail to live) is y^ x H = 1^2; the probability that the man will live and his wife will fail to live is % X H = %; the probability that the wife will live and the man will fail to live is H X % = H- Notice that the sum of these four probabilities is%+K2+14+%= ' 1. One of the four cases must happen. 18 MATHEMATICS OF LIFE INSURANCE [§17 Exercises 1. A single die is thrown twice in succession. What is the prob- ability of getting 3 the first time and 5 the second time? If two dice are thrown at once, what is the probability of getting a count of 8? Of getting a count greater than 9? 2. One bag contains six white balls and five red balls, and a second bag contains four white balls and three black balls. A ball is drawn from each. What is the probability that both are white? That one is white and one is black? That one is white and one is red? That one is red and one is black? 3. What is the probability of holding four aces in a game of bridge? 4. A, B, and 0 shoot at a mark. A's record is one hit in three, B's record is two hits in five and C's record is three hits in seven. Tabulate the eight possible contingencies that may arise and deter- mine the probability of each. B. In the preceding exercise, the shots are taken in the order A, B, C.The first person to hit the mark wins the match. Determine the probability of each to win the match. 6. In a certain game with dice, it was found that 4871 games out of 9900 were won by the person throwing against the house. What is the probability that any one person will win against the house? 17. Total Probability.—Events are said to be mutually exclusive if the happening of any one of them excludes the happening of any of the others. Thus, if A, B, and C engage in a game in which there can be but one winner, the events "Awins," "B wins," "C wins" are mutually exclusive. The probability that some event will happen out of a series of n mutually exclusive events is called total probability. Notice that, if any event of the series happens, the rest must necessarily fail to happen. The probability that a particular event of the series happens when taken by itself is called partial probability. The determination of total probability when the several partial probabilities are known rests upon the following theorem: Theorem.—The total probability that some event will happen out of a series of n mutually exclusive events is the sum of the n partial probabilities that each event mil happen when taken by itself. }17] THE THEORY OF PROBABILITY 19 If, as before, 0:1, a-s, 03, ... a.n indicate the number of yays the respective events can happen when each is taken by tself, then the number of ways favorable to the occurrence of lome event is a-i + 02 + as + . . . + a.n. If m represents the .otal number of possibilities, favorable and unfavorable, then = al Jr al ~^~ a3 ~^ a" = al 4- a2 -i- a3 4- 4-"" — m ~ m m m m = PI+PS+PS+ + Pn, (1) vhere, again, the p's represent the respective probabilities that each 'vent will' happen when taken by itself. Problems involving total probability are indicated by the word 'or." For example, if the probability A, can win a race is % and he probability B can win it is ^i, then when they run with other iompetitors (assuming no "ties") the events "A wins," "Bwins," ire mutually exclusive. The probability that A wins or B wins s % + % = %. Again, suppose the probability that A can solve a problem is me in three and that B can solve it is two in three. The events 'A solves it," "B solves it" are not mutually exclusive, since hey may both solve it. The mutually exclusive events are, 1. Both solve the problem, probability % X % = %. 2. A solves it and B fails, probability % X % = %. 3. A fails and B solves it, probability % X % = %. 4. Both fail of solution, probability % X % = %. Tie problem will be solved in any one of the first three cases; hat is, if both solve it or A solves it and B fails or B solves it and L fails. Hence, the probability the problem will be solved when oth work at it is %+%+%=%. Or one may reason thus: ^he problem will be solved if they do not both fail of solution. lence, the probability the problem will be solved is 1 — % = %. Exercises 1. A, B, and C run in a race with other competitors. The prob- bility that A will win is one in five, that B will win is one in four, nd that C will win is one in three. What is the probability that one f the three will win? 20MATHEMATICS OP LIFE INSURANCE [§18 2. The probability that A can solve a certain problem when work- ing alone is one in three and the probability that B can solve it is one in four. What ,is the probability that the problem will be solved when both work at it, assuming they do not assist each other? 3. An um contains five black balls, three red balls, and two white balls. If three balls are drawn from the urn, what different com- binations may result, and what is the probability of each? 4. In a single throw with a pair of dice what is the probability that neither ace nor doublets will appear? 6. From a lottery of thirty tickets, marked 1, 2, . . . , 30, four tickets are drawn. What is the probability that Nos. 1 and 15 will be among them? 6. In a single throw, what are the relative chances of throwing a count of 9 with two dice and with three dice? 7. In two trials with a single die what is the probability of throwing an ace the first time only? At least one ace? 18. Repeated Trials.—When the probability that an event will occur in a single trial is known, it becomes a question of importance to determine the probability the event will happen a specified number of times in a given number of trials. For example, sup- pose a die is thrown 6 times in succession, or 6 dice are tossed at once, and we desire to know the probability of getting exactly one ace; exactly two aces; and so on to exactly six aces; or no aces. Such questions are answered on the basis of the preceding sections. The probability that a particular die falls ace upper- most and the rest fail to do so is a special case of Sec. 16, consequence 4; that is, it is % (%)6. But to obtain exactly one ace it is only necessary that any one of the six dice falls ace uppermost and the rest fail to do so. Hence the probability of getting exactly one ace is the total probability 6 X %(?^)6 = (%)5 = 312^g. The probability that some particular two of the dice fall ace uppermost (and the rest fail to do so) is y^ (%)4. But there are sCi = 15 possible groups (combinations) of two dice in six; and hence the total probability of getting exactly two aces is MWHV = 15 %e "^oe = 3125/15,552. The probability that all six of the dice fall ace uppermost is (%)', and the probability that no ace is obtained is (%)6. §181 THE THEORY OP PROBABILITY 21 Collecting the above results, we see easily that the probabilities of getting no ace, one, two, three, four, five, or six aces are given by the terms of the binomial expansion (i + i)6 = O)6 + 6ft)0)° + l5(i)»0)« + 20(i)3^)' + 15(^0)' + ea)5® + (i)6. The fourth term of this expansion is the probability that three of the dice fall ace uppermost and the other three fail to do so. In general, we have the following theorem: Theorem.—If p is the probability that an event will happen in a single trial and q is the probability that the event will fail to happen in a single trial, then the probability that the event will happen exactly r times in n trials is -c'w-' = i^ryr-. (D Imagine the n trials grouped in combinations of r trials each. The probability that the event in question will occur in all the trials of some one of these groups and fail to occur in the remaining n—r trials is the partial probability py. But there are nCr such groups, and hence the probability that the event will happen exactly r times in the n trials is the total probability nC'rp'y-'-. We may modify the questions at the beginning of this section and inquire: What is the probability of getting at least one ace? At least two aces, and so on? The probability of getting at least one ace in six trials with a single die is the sum of the partial prob- abilities of getting exactly six, or five, or four, or three, or two, or one aces, and hence is given by the sum of the fractions ft)6 + 6(mf) + ISft)4^)2 + 20(i)3?)' +15 ft)'0)< + 6 (i)W. Instead of computing the values of these fractions and adding, it is more convenient to use consequence 3 (Sec. 16). Thus, we shall get at least one ace if we do not fail every time. The probability of failing every time is (%)6 = 15,625/46,656. Hence, the proba- bility of getting at least one ace is 1 - (%)< = 31,031/46,656. The probability of getting at least two aces in six trials with a single die is the sum of the partial probabilities of getting exactly 22 MATHEMATICS OF LIFE INSURANCE [§18 six, or five, or four, or three, or two aces and hence is given by the sum of the fractions ^)6 4. ^)5^-) + l5(i)<(^ + 20a)3(l)3 + 15(i)2^)4. But the sum of these fractions is the sum of all the terms in the expansion of (H + ^"except the terms 6(%)(%)° + (%)''. Since the sum of all the terms in (H + W is I6 = 1, the probability of at least two aces in six trials is 1 - [6(^)(%)5 + (%)6] = 8531/46,656. Theorem.—Ifp is the probability that an event will happen in a single trial, and q is the probability the event will fail to happen in a single trial, then the probability that the event will happen at least r times in n trials is either (=n t=r-l YnC'.pT-'orl - ^ nC.p'g"-'. (2) (-r <-0 Expanding the first of these expressions by giving to ( all inte- gral values from r to TO inclusive and adding, we get .Crpy + nC'r+lP^'ff"-'-1 + nCr+lP^-'-'1 + . . . + P"; (3) that is, the total probability that the event happens exactly r times, or exactly r + 1 times, or exactly r + 2 times, ... to exactly n times, or the probability the event happens at least r times in the n trials. Expanding the sum in the second expression, we find qn+^p<r~l+nC^q''-'t+.C^psqn-r+ . . . +nC',-lpr-ln^n-T+l; (4) that is, the total probability that the event fails to happen in any of the n trials, or happens exactly once, or exactly twice, ... to exactly r — 1 times, or the probability that the event happens at most r - 1 times in the n trials. Subtracting this probability from 1 gives the probability that the event will occur at least r times in the n trials. Notice that the sum of expressions (3) and (4) contains all the terms in the expansion of (p + g)". In practice one uses that one of the expressions in the theorem which requires the least amount of labor to compute. §18) THE THEORY OF PROBABILITY . 23 Another problem of some interest in connection with repeated trials is the following: In how many trials will the probability that an event happens at least once be a given proper fraction A? If q is the probability that the event fails to happen in any one trial, the probability that it will happen at least once in n trials is 1 — q\ We have, then, to solve the equation
for n. Hence,
n =
1 - ff" = fc log (1 - k)
logq
For example, In how many trials with a single die will it be an even chance of getting one ace? Here k = % and q = %. Hence, logi -0.30103 „ " = log4 ° -0.07918 - -8- Since n is necessarily an integer, the probability of getting at least one ace in four trials is greater than one in two, and the probability of getting at least one ace in three trials is less than one in two. In fact, the two probabilities are, respectively, ,y^ = 0.517, nearly, and AS- = 0.421, nearly. Exercises 1. Find the probability of throwing exactly three aces in five trials with a single die. At least three. 2. Find the probability of throwing doublets one or more times in three trials with a pair of dice. 3. A certain stake is to be won by the first person who throws 5 with a die of twelve faces. What is the chance of the sixth person? 4. A and B play chess. A wins on the average two games out of three. What is A's chance of winning just four games out of the first six? 6. In how many trials will the probability of throwing double 5 with a pair of dice be four in seven? 6. Find the probability of throwing at least four aces in six trails with a single die. 24MATHEMATICS OF LIFE INSURANCE [§19 7. Onan average seven ships out of eight return to port. Find the probability that out of five ships expected at least three will return. 8. I have two purses. In the first are nine nickels and one five- dollar gold piece and in the second are ten nickels. Nine coins are taken from the first and placed in the second, and then nine coins are taken from the second and replaced in the first. What is the proba- bility that the gold piece is in the first? 19. Expectation in n Trials.—If a single die is thrown 12 times in succession, or twelve dice are thrown at once, we intuitively expect two aces will appear. Our intuition is based upon the fact that the probability of getting exactly two aces is greater than the probability of getting any other number of aces. That is, the term ^C^^)2^)10 is greater than any other term in the expansion of (H + %)12. To say that an event is expected to happen r times in n trials means that the probability that it will happen exactly r times in the n trials is greater than the probability that it will happen exactly any other number of times (or at least as great). The expansion of (p + ?)" may have one term greater than any of the others, or it may have two terms of equal magni- tude greater than any of the others. For example, the greatest term in (H + %)18 is uCa OO2^)11. K a die is thrown 13 times, we should expect two aces. But the expansion of (% + %V has two terms of equal magnitude which are greater than any of the others, namely, nC'2(i)^)15 and irCad)3^14. Hence, if a die is thrown 17 times, we should expect either two aces or three aces. If the term nCrpy is the greatest term in the expansion of (p -(- g)n^ the ratios of the two consecutive terms to it are each less than 1. Hence, we must have (Sec. 13,Eqs. (4) and (5)). —————.?<1 and n-1-p<l. (1) n — r + 1 p r + 1 q Clearing these inequalities of fractions, qr < pn — pr + p and pn — pr < gr + q. (2) }19] THE THEORY OF PROBABILITY 25 Transposing and remembering that p + q = 1, r <pn + p and pn — q < r. (3) Combining these two inequalities, we have finally, pn + P > r > pn — q. (4) But if nCrp'q"^ is equal to either of its consecutive terms, the signs of inequality in the above analysis are replaced by signs of equality and the word "and" by the word "or." We then find in place of (3) r = pn + p or pn — q = r. (5) Since r is necessarily an integer, neither of Eqs. (5) can hold unless pn + p[ = p(n + 1)] is an integer; and then pn — q [= pn — (1 — p) = p(n + 1) — 1] is also an integer. The expansion of (p + q)" then has two terms of equal magnitude greater than any other term. Thus, if seventeen dice are thrown, pn + p = 1^ + Vv, = 3 and pn - q = 1^ - % = 2 and r may be either 2 or 3. If pn + p is not an integer, neither is pn — q an integer, and the inequalities in (4) must hold. Since (pn + p) — (pn — q) = p + q = 1, r is the integer lying between pn + p and pn— q. Thus, if fourteen dice are thrown, pn + p = ^ + % = 1^ = 2% and pw — q == 1^ — % = % = 1H. Here r is the integer 2 lying between the fractions 1% and 2^. From (4) we easily derive ^4>^>.-h- (6)
n + 1 "' n + 1 w The two limits between which p lies differ by r+1 _ r 1 n+1 ra+l'n+l' which approaches zero as n increases. Hence, the two fractions have a common limit. This common limit is seen to be r/n by writing (6) in the form r.+i n——'->?>———
^ -^ 26 MATHEMATICS OF LIFE INSURANCE [§19 Hence, 7-n^ (7) which is Bernoulli's theorem (Sec. 15). As an example, the probability that a child of age ten will live to age fifty lies between the fractions 69,805 n.oon/iQ j 69,804 faenw 100,001 = 0-698043 and 100,001 = 0-698033' since, here, r = 69,804 and n = 100,000. Hence p = 0.69804 with an error of less than 1 in 100,000. Bxercises 1. A single die is thrown 12 times in succession. How many aces will be expected? What is the probability of obtaining exactly as many aces as expected? 2. Write the arithmetical expressions for the probabilities of obtaining exactly zero, exactly one, exactly two, and so on to exactly twelve aces in twelve throws with a single die. Choose a convenient scale and plot these probabilities against the corresponding number of aces. Draw a smooth curve through the plotted points. Such a curve is a probability curve. 3. A life insurance company has issued exactly 5000 policies to as many persons of age x. The death rate between age x and age x + 1 is known from experience to be one person in every fifty. How many policies will the company expect to pay at the end of the first year? Write the expression for the probability of paying exactly as many of these policies as expected on the basis of the first theorem in Sec. 18. 4. An approximate expression for the probability of expectation, when the number of trials is great, is given by the expression V'2impg where n is the number of trials, p is the probability the event will occur in any one trial, q is the probability the event will fail in any one trial, and v = 3.1416. Making use of this formula, compute the probability of expectation in Exercise 1 and compare with the result there found. Compute the probability of expectation in Exercise 3.