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CHAPTER XVIII

PROBABILITY

1. Suppose that a bag contains a hundred balls of exactly the same size and shape, of which one is coloured and the remaining ninety-nine white. If one of these balls be drawn at random it is safe to say that there is a greater probability of drawing a white ball than of drawing the coloured ball. Again, of a number of men in a community, all subject to the same conditions, the probability that one aged 20 will sur^vive 10 years is obviously greater than the probability that one aged 90 will survive the same period. These examples serve to illustrate what is understood by the term " probaMany more such simple examples could be given, from which we could say that the probability that event A would happen is greater or less than that event B would happen. A difficulty arises, however, when we attempt to assign a measure to the probability that one of the events A or B might happen.

In the first of the above examples it is not unreasonable to argue that since there is one coloured ball among a hundred balls, the probability that this coloured ball would be drawn is one in a hundred. Whether this be so or not, we have used the data at our command and have given a measure to the probability. In the second example, however, we have no data immediately available to enable us to assign a numerical value to the probability in question. It would be necessary to collect statistics of the mortality of the men in the community, and to supplement the collection of these statistics by more or less elaborate calculations. In the application of the theory of probability to most actuarial problems the aggregation of statistics is necessary before the data are avail-able, and it is not often that we are immediately in possession of such simple facts as are given in the first type of question. A comanswer to a problem in -actuarial work would involve the collection and interpretation of the relevant statistics : these are matters which are outside the scope of this book, and in dealing with questions on probability it will be sufficient to assume that

314 PROBABILITY
we have sufficient data to enable us to proceed without further analysis.
2. Numerical definitions of probability.
A simple definition of the measure of the probability that an event may happen is as follows :

(i) If an event can happen in a ways and fail to happen in b ways, all of these ways being equally likely and such that not more than one of them can occur, then the probability of the event happening is —

a
a+b
The probability that the event does not happen is _a^b_b, which
a
is equal to 1 —

a b'
Although this definition enables us to solve arithmetical problems in probability, it has, however, certain defects which render it unsatisin general. In the first place we have to define the phrase "all of which are equally likely." It is difficult to say what this means without rendering the definition unintelligible, for if we define the words as equivalent to "there being no reason why one way should occur more than another^" we should be involved in a further definition. Moreover, however we explain the phrase it may not be possible to say that the happenings of an event are equally likely: the probability of its occur

rence cannot then be measured by the simple ratio a + _b. The attempt

that has been made to define "equally likely" events as those which occur equally often in the long run, i.e. when the number of occasions is very large, does not seem to solve the dif^ficulty.

While the above definition—which is sometimes called the "unitary" definition—is sufficient for solving the questions inprobability that will be dealt with in this chapter, it will not be out of place to quote other definitions.

If on taking any very large number N out of a series of cases in which an event A is in question, A happens on pN occathe probability of the event A is said to be p. (Chrystal, Algebra, chap. xxxvi.)

If there be a large number N of a series in which an event M might occur and if, whenever it might occur it would necessarily

NUMERICAL DEFINITIONS OF PROBABILITY 315
be associated with one of two or more mutually exclusive events E, E', E^", ... , then if E would occur on pN occasions, E^' on p' N occasions, ... and so on, p, p^', p^", ... are said to be the probaof the event M being associated with E, E', E", ... , and p + p^'+ p^"+ ... = 1. (Sheppard, Ency. Brit., ^"Probability.^")
The term" mutually exclusive "in the above definition means that on each occasion one and only one of the events E, E', E", ... can occur.
We can arrive at an approximate agreement between these deand the unitary definition thus. If an event occur on a occasions out of (a + b), we may say that the frequency of its occuris a, and the relative frequency will be a/(a + b). Then the probability of the event is the relative frequency when the number of occasions is very large.
For example, we may revert to an illustration similar to the first illustration in paragraph 1. If there be five exactly similar balls in a bag, two coloured and three white, then by the unitary definition the probability that a coloured ball will be drawn (supposing one ball be drawn at random) is 5. This does not mean that if a ball were drawn five times, being replaced each time, two of the draws would necessarily give coloured balls. The true interpretaof the fraction 4 is that if the experiment were repeated a large number of times a coloured ball would be drawn about twice in five times. After many trials we should find that twice in five times was approximately correct, and we might be said to have confirmed our unitary result by experiment.
There are thus two distinct methods of ascertaining simple probabilities, one from a priori considerations, and the other by experiment. Usually only one of these methods is applicable to any particular problem, and for the present purpose it will be unto consider statistical methods.
3. The basis for the solution of the problems that will be dealt with later is, as has been remarked, the unitary definition. We may amplify this definition by certain others.
(a) The measure of certainty is unity. In other words, if an event is certain to happen the probability of its happening is 1 ; if it is certain not to happen the probability of its happening is o.

316 PROBABILITY
(b) If p is the probability that an event might happen, the probability of its failing is 1 — p. This follows directly from the unitary definition.
(c) If the odds in favour of an event are a : b, the probability of its happening is a'(a + b).
(d) If an event can happen in more than one way, all ways being mutually exclusive, the probability of its hapat all is the sum of the probabilities of its happening in the several ways.
This last proposition is known as the ^"addition^" rule, and is fundamental in the application of the theory of probability. The rule may otherwise be stated thus:
If p and p^' are the probabilities of two mutually exclusive events, the probability that one of these events happens on any occasion in which both events are in question is p + p^'.
The words "mutually exclusive" in this definition are of the utmost importance. A proof of the addition rule depending on Chrystal's definition of probability will show that unless the events are mutually exclusive we cannot add the simple probabilities p and p'. Thus, if there be a large number N out of a series in which the two events are in question, the first will happen on pN occaand the second on p'N occasions respectively. Since the events are mutually exclusive, one and only one of the events can occur on any one occasion. Therefore out of N occasions one or other of the events will happen on pN + p^'N occasions, i.e. on (p + p^') N occasions. The probability that one of the events hap-pens is therefore p + p^'. It will be seen that this proof breaks down if the second event happens on any of the pN occasions that the first happens, i.e. if the events are not mutually exclusive.

4. The simplest problems in probability are those in which both the number of favourable ways and the total number of ways in which the event may happen can be counted, either arithmetically or by the aid of elementary rules. The enumeration is often best performed by the application of the theorems of permutations and combinations, and a thorough knowledge of this branch of algebra is essential for the speedy solution of many of these questions.

ILLUSTRATIVE EXAMPLES Example 1.

An ordinary pack of cards contains 13 cards of each of four suits; spades, hearts, diamonds and clubs. Ten cards of each suit are numbered i (the ace), 2, 3, ... 10, the remaining three being ^"court cards,^" namely, king, queen, knave.

Find the chance that if a card be drawn at random from an ordinary pack it will be a heart.

Since there are 52 cards in a pack, the number of ways in which a card can be drawn (i.e. the total number of ways in which the event can happen) is 52.

The number of ways favourable to the event is 13, since there are 13 hearts in the pack.
The required probability is therefore 1'z or } .

Notes: (i) The above result means that over a long series of trials the proportion of favourable draws would approximate to one-fourth of the whole number of trials. It is convenient, however, to think in terms of a single trial, although the results of our calculations merely epitomize the experience of a long series of trials.
(ii) The word ^"chance " may be taken to be the same as the word "probability." There is a shade of difference between the two ideas, ^"probability^" referring to past, present and future events, and "chance" to future events only. This distinction can be ignored.

Example 2.

In a given race, the odds against three horses A, B, C are 2 : I ; 5 : 2; to : i respectively. Assuming that a dead-heat is impossible, find the chance that one of them will win the race.

The events are mutually exclusive, and we may therefore use the "addition" rule.
p^'= probability that A wins = p„= B ,
^p,_
,5 ^C73
The probability that one of the horses wins is therefore
_p, + _p„ + _p,,, = ¹~
Example 3.

Out of 211 + I tickets consecutively numbered, three are drawn at random. Find the chance that the numbers on them are in arithmetical progression.

318 PROBABILITY
This problem can be sol^ved by virtually writing down the sets of numbers that can be in A.P. Consider, for example, the sets of three numbers in A.P. beginning with 4: they will be
45;⁶ 4;⁶;⁸ 4;7;10 ...4;71+I;2n—2 4;n+2;211,
where the highest number that can enter into any set is 2n. There will evidently be n — 2 such sets of numbers (being the number of numbers from 5 to n + 2, both numbers inclusive). We can therefore write down the following schedule :

Lowest number of the three	Number of favourable ways
I	n
2	n — I
3	71 — I
4	n—2
5	n — 2
6	71—3
2n—2	I
*271—* I	I

The total number of favourable ways is 2n (n + ^I) — nor n⁼, and since 2

the total number of ways of drawing three tickets is (2n I) 2n (211 — I) 6
it is easily seen that the required probability is 3⁷¹471— I Example 4.

If the letters of the word REGULATIONS be arranged at random, what is the chance that there will be exactly four letters between the R and the E?

The conditions are satisfied if
the R is in the first place and the E in the sixth,

or „ R		second „	E	„	seventh,
„	R	third „	E	,,	eighth,
„ „	R	fourth „	E	„	ninth,
„	R „	fifth „	E	„	tenth,
„	R	sixth	E	„	eleventh,

or if the R and the E are transposed in any of the above.

THE MULTIPLICATION RULE 319

Now there are ii letters in the word; therefore the letters R and E can jointly occupy I I x to positions. Since there are 6 x 2 favourable positions (as above) the chance that the R and the E are in the favourable position is _Tyr or A.

Example 5.
Find the probability that the number 8 will be thrown in a single throw with two dice.

The total number of different numbers that can be thrown with two dice is 36, since each of the 6 ways of throwing the first die can be associated with each of the 6 ways of throwing the second.

The number of ways favourable to the event can be found by simple enumeration, thus:

First die	Second die
2	6
3	5
4	4
5	3
6	2

Total: 5 ways. The required probability is therefore A.

5. The multiplication rule.
If there be a series of events such that the happening of any one of them in no way affects the happening of any other of them, the events are said to be independent.
The probability that two independent events happen on any one occasion on which they are both in question is the product of the chances of their happening severally.
For consider two independent events C and D, the probabilities of which are p and p^' respectively. Out of a large number (N) in which the events are in question C will happen on pN occasions, and out of these pN occasions D will happen on p^' (pN) or pp^'N occasions. The probability that they both happen is therefore pp^'.
This rule is known as the ^"multiplication^" rule, and its applicato the probabilities of happening or otherwise of two indeevents may be seen from a consideration of the following schedule.

Let the probabilities of happening of two independent events be p and p^' respectively. Then

3 20 PROBABILITY

(i)the chance that they both happen = pp^';

the chance that the first happens and the second fails

^=p

^—pt

);

(iii)the chance that the second happens and the first fails

=p'(I —p);

(iv)the chance that they both fail = (i — p) (i — p'). The total of these chances

=pp'+p(I —p')+p'(I —p)+(I — ^p)(^I^—P^')

=pp'+p—pp'+p,—pp'+

^—p—p'

^pp'

= I,

which is obviously true, since we have exhausted all the possibilities.

If the two events are not independent, we may still apply the multiplication rule. Let'', be the chance of happening of an event, and p₂ the chance of happening of a second when the first has happened. The chance that they both should happen is p, p,.Care must be taken to distinguish between independent and dependent events. In considering independent events the hapof either event does not affect the happening of the other, and to find the probability that they both happen we multiply the simple probabilities of each event. Where the events are not independent, we must find the chance p₂ that the second happens when the first has happened before we can apply the multiplication rule. For example, if there be two urns, one containing one white and two coloured balls, and the other one white and three coloured balls, the probability that the combined result of two drawings, one from each urn, will give the two white balls is ₃x 4, since the drawing from the first urn in no way affects the drawing from the second. If, however, we were required to find the chance that in two successive drawings from an urn containing two white and five coloured balls the two white balls would be drawn (a ball drawn not being replaced) we should reason thus. The chance that a white ball is drawn at the first drawing is ; after this has happened there will be left one white and five coloured balls, and the chance of drawing the other white ball at the second drawing will be '~. The combined chance is therefore x

ILLUSTRATIVE EXAMPLES 32I

The multiplication rule can be applied to the probability of happening of any number of independent events. If p, p^', p^", ... be the chances of the events happening severally, the chance that they all happen is pp'p^" ... .

Again, if the chance that an event happens in a single trial be p, the chance that it will happen in each of

trials is

and the chance that it will fail in each of, say,

(n —

trials is

— p)n-r

Suppose that it is desired to find the chance that an event will happen

exactly r

times in

trials.

Let the chance that in any one trial the event does not happen be

I —

p).

Then if the event happen in

trials and fail in

n — r,

the combined chance is

ⁿ

But of the

trials we can select the

trials in which the event happens in

n,.

ways. The chance that the event happens in exactly

trials out of the

n is

the chance that it happens in these r trials and fails in the remaining

n —

trials. The required chance is therefore

nrp

ⁿ

Now it will be noted that

ⁿ

is the term containing

in the expansion of (p

+ q)

ⁿ

The successive terms of this expansion will therefore give the probabilities of the event happening

n, n — I,

n —

2, ...

n — r, ... times exactly inn trials.

The application of the above principles can best be appreciated by reference to actual examples.

Example 6.

Find the chance that (i) three heads, (ii) two heads and one tail will

turn up in three successive spins of a coin.

(i) The chance of a head at the first spin =

„second „

= 2

third>> = p

ⁱ

Total chance = (1)3 =

(ii) The chance of a tail is the same as that of a head, so that the

total chance might appear to be

(2)

as before. This is in fact the chance

that

heads and

tail occur in a specified order, say, HTH. The conof the question would, however, be satisfied if the order were HHT or THH. These three favourable events are mutually exclusive.

The chance that the two heads and tail are obtained in one of the three

orders is found by adding the three probabilities. The required chance

is therefore 3 x

(1)

= -

322 PROBABILITY
Example 7.

Find the chance of drawing a king, a queen and a knave in that order from an ordinary pack in three consecutive draws, the cards drawn not being replaced.

This is an example of the application of the multiplication rule to dependent probabilities.
If p_i be the chance of drawing a king;
p₂a queen, a king having been drawn;
p₃ a knave, a king and a queen having been drawn,
then the chance of drawing a king, a queen and a knave in succession
is ^pi^p2^p3

Now there are 4 kings in the pack, so that the chance of drawing a king is When the king has been drawn there are 51 cards left of which 4 are queens. The chance of drawing a queen is therefore 5 . Similarly, the chance of drawing a knave from the remaining 5o cards is

P1 = ₅2; P2⁼ r_'P3= ~1^.
3

The required chance = ^pl^p2^p3 = 52 5 3 5° .
Example 8.

A bag contains three balls, one red, one white and one blue. X and Y draw a ball at random alternately. If X draws the red ball or Y the white ball it is retained. Otherwise the ball drawn is immediately replaced. Find the chance that just before the fifth draw is made the blue ball only is in the bag.
Let R, W, B denote the red, white and blue ball respectively. Now the blue ball is the only one in the bag just before the fifth draw if any of the following have happened :

	X	Y	X	Y
	first draw	first draw	second draw	second draw
(i)	R	B	B	W
(ii)	R	B	W	IV
(iii)	B	W	R	B
(iv)	W	W	R	B
(v)	R	W	B	B
(vi)	B	R	R	W
(vii)	B	B	R	W
(viii)	W	R	R	W
(ix)	W	B	R	W

ILLUSTRATIVE EXAMPLES    323
The chances of these events happening are:    Product
    (¹)    2    2    2

₍l_am

1¹¹)    3    2    2    2
(iii)
3    2    18
(iv)
3    2    ₁1_₇

(v)¹IS
(vi)33325¹4

(vii)

(viii)

(ix)

The total chance is therefore

1,:

+ 4

+ 1

+= >

Note. A systematic enumeration of possibilities is of great importance, and although in the above example the work may be shortened by alternative methods, for illustrative purposes the eventualities have been set out in full.

Example 9.

A bag contains three red and three green balls and a person draws out three at random. He then drops three blue balls into the bag and again draws out three at random. Show that he may just lay 8 to 3 with advantage to himself against the latter three balls being all of different colours.

After the insertion of three blue balls the bag may contain:

	Red	Green	Blue
(^a)	3	—	3
(b)	2	I	3
(^c)	^I	²	3
(^d)	—	3	3

The probability that the bag contains (a) is -. 6 . or for the chance that a green ball is drawn is obviously , the chance that a second green ball is drawn is and the chance of a third green ball is then l . Similarly,

the chances under (b), (c), (d) are o, A, respectively.
Nov three different colours on the second draw can only be obtained if the six balls come under headings (b) and (c).

Under (b) the probability at the second draw of drawing three different coloured balls is A, and therefore the compound probability that this will happen is A. ,r. Under (c) the probability is the same.

The total chance is the sum of these and amounts to Therefore,
since the odds against the favourable happening are 73 to 27, the person
2I-2

324 PROBABILITY
drawing the balls may lay 8 to 3 against his drawing three of different colours and obtain a slight advantage.
Example 10.

Q is the probability that a man aged x will die in a year. Find the probability that out of five men, A, B, C, D, E, each aged x, A will die in the year and be the first to die.

The chance that a given man dies in the year = Q.

The chance that a given man does not die in the year =	1 — Q.
The chance that none of the five dies in the year =	(1 — Q)⁵.
The chance that at least one man dies in the year =	— (1 — Q)⁵.

Since the chance that A is the first to die is obviously , and since this is independent of the chance that at least one man dies in the year, the required chance = {1 — (i — Q)⁵}.
8. The methods for the solution of the above examples depend largely on the simple application of the formulae for permutations and combinations. It is safer as a general rule to use permutations rather than combinations. If repetitions are not allowed this is not very material, as each combination of r things forms r! permutaIf, however, repetitions are allowed then in most instances it is essential to use arrangements.
Other algebraic devices are often useful. For example, in dealing with questions involving the sum of the numbers that can be thrown with dice and kindred problems dealing with homoneous products, it is of advantage to employ the binomial theorem.
The following examples illustrate the use of this method.
Example 11.
Four dice are thrown. Find the chance that the sum of the numbers appearing will be 18.
Regard being had to the different ways of making up the same total, the number of numbers that can be thrown with four dice is the sum of the coefficients in the expansion

(x + x` + x³ + x4 + x5 + x6)4_.
The sum of the coefficients will be found by putting x = I. The total number of possible numbers is therefore 6⁴.
The number of ways of throwing 18 is the coefficient of ^xisin the above expansion.

ILLUSTRATIVE EXAMPLES    3²5 The coefficient of x¹⁸in (x + x² + x³+ x⁴ + x⁵+ _x6)4
x¹⁸inx⁴(I +x+x²+x³+x⁴+x⁵)⁴
x¹⁴in (I + x + x² + x³+ x⁴ + x⁵)⁴x¹⁴In ⁽¹-    ^x6)4
(I — x)⁴
x¹⁴In (I — x⁶)⁴(1 — x)^-4
x¹⁴in(I -4x⁶+6x¹²...) (I+4x+ lox' +...
9.10.II ₈15.16.17 14
+ ₆    x +...+    ₆x" +
— 15.16.17 — 4.9.10.11
6    ₆    + 6. lo = Ho.
The required chance is therefore ₆80

Note. We are here dealing with arrangements which would be disif one die were red, one blue, one yellow and one green. Selections in an example of this type would give quite wrong results.

Example 12.

Nine cards are drawn at random from a set of cards. Each card is marked with one of the numbers 1, o or — 1, and it is equally likely that any of the threg numbers will be drawn. Find the chance that the sum of the numbers on the cards thus drawn is zero.
The number of favourable drawings will be the coefficient of x° in the expansion of (x^-1 + x° + x¹)⁹; the total number of possible drawings will be the sum of all.the coefficients in the same expansion.

The coefficient of x° in (x^-1 + x° + x')9
x° in x-⁹ (1 + x + x²)⁹= x⁹ in (1+x+x²)⁹
x⁹ in (I — x³)⁹(I — x)-⁹.
Proceeding on the same lines as in Example II, the required proba
bility is found to be ³¹9⁹. 3

9. The theory of probability was evolved from a consideration of games of chance, and many problems dealing with these games can be solved by the elementary methods outlined above. No new

,,

326 PROBABILITY

principles are involved; all that is required in attacking these problems is a clear understanding of the particular happenings that may arise. The following examples are illustrative of the methods : the first of these examples is analysed fully, and the student should be able to solve problems of the same type by the application of similar reasoning.

Example 13.
A and B throw alternately with a pair of ordinary dice. A wins if he throws six before B throws seven, and B if he throws seven before
A throws six. If A begins, show that his chance of winning is R. (Huyghens^' Problem.)

First, let us consider the chances of throwing six or seven with two dice. We can find the number of ways of throwing these numbers either by actually counting the number of ways (as in Example 5) or by finding the coefficients of x⁶ and x⁷ in the expansion of (x + x² + x³ + x⁴ + _x5 + _x6)2 (as in Example I I). Since the figures are small, the first way is simpler, and it is easily seen that six can be thrown in 5 ways, and seven in 6 ways. The chance of throwing six in one throw with the two dice is therefore

and of throwing seven A.
A can win if he throws six the first time. His chance of throwing this number is A.
He may fail to throw six the first time. He can then win if B fails to throw seven at his first throw and A throws six with his second throw. The chance that A fails to throw six is I —    = N.
The chance that B throws seven is    and the chance that he fails
to throw this number is I    =

Therefore the chance that A wins at the second throw is R.N.-A.

If A fails to win at the second throw, he can win at the third throw if
B has not thrown seven at his second throw. The chance of this is
s1 30 :'1 ro and so on.
A^'s chance of winning in the long run, i.e. after a very large number of trials, is therefore
5 31 30 5 31 :30 31 30 5

_ S (I + r + r² + r³+ ...),

where r =
3a•:;c•
If we sum this geometrical progression to infinity, we find that A's chance

_ 5 I _ 5 1296 30 36 1 — r 36' 366 61 '

ILLUSTRATIVE EXAMPLES 327 Example 14.

A and B play a match to be decided as soon as either has won two games. The chance of either winning a game is 1-, and of its being drawn ₁"_u. What is the chance that the match is finished in lo or less games?

If the match is not finished in lo games, the following must occur:

(i)All the games must be drawn ; or

(ii)

must win one game and the remaining

₉

must be drawn ; or

(iii)

and

must each win one game and the remaining 8 must be drawn.

The chances of these mutually exclusive events are:

₍₉₎

(i)

`I0

since this result may occur in to different ways and either A or B may win ;
since the number of orders in which this may occur is ¹⁰P₂
The chance that the match is not finished in to games is, by the addition rule,
Example 15.

A, B, C, D each throw two dice for a prize. The highest throw wins, but if equal highest throws are made by two or more players, those players continue. A throws q, B throws 7. Find C's chance of winning the prize.

Consider the following scheme, showing the total number of possible ways in which C may win:
2. Io^th'

i.e. the chance that the match is finished in to games or less

9^8.3⁸7
= I — 22.'13010 = • 17 approximately.

328	PROBABILITY
328	PROBABILITY
C may win if		Probability
C throws Iz, D less than 12		^I	35
		6²	36
C throws 12, D 12, and C wins later		I 6²	I 6²	I* 2
C throws I1, D less than I1		²	33
		6²	36
		2	2	1*
C throws I1, D It, and C wins later		6	6²	2
C throws to, D less than to		3	3⁰
		6²	36
C throws lo, D Io, and C wins later		3	3	^I*
		6²'	6²	2
		4	26	I*
C throws 9, D less than 9 and C wins later		6²	36	2
(since A has thrown 9) C throws 9, D 9, and C wins later		4	4	1*
		6²'	6²	3

The total of these chances is ;T.
10. Most probable value.

It is essential to distinguish between the absolute probability of an event, the average number of times that it may happen over a series of trials and the most probable number of times that it will occur. The most probable value is the value that occurs with greatest frequency, and in simple examples this value can be easily deter-mined by considering separately the contingency of each event. Thus, in a single throw with two dice the numbers that may turn up and the chances of occurrence of these throws are given in the following table :

Possible numbers: ²3 4 5 ⁶7 ⁸9 10 1 1 12 Chances of occurrence : 1 2 3 4 5 ⁶5 4 3 2 I (- 36)
The most probable number to be thrown is therefore 7.

Again, in Example 17 below, a hasty conclusion would be that half the pack (i.e. 26 cards) would have to be turned up before two aces out of the four would appear. This is not so, however.

* The final factor in each of these terms expresses the chance that C will
win after having equalled the number thrown by D and/or A.

MOST PROBABLE VALUE 329

The four aces divide the remaining 48 cards into 5 groups. If a large number of such divisions were made the average number in each of the 5 groups would be 48/5. The average number of cards which must be turned up before two aces appear is therefore 2 of these groups plus 2 aces, i.e. 96/5 + 2 or zi•z, as given below. The solution on p. 330 shows that z6 is neither the average number nor the most probable number, and, further, that there is a definite probability associated with each number of cards turned up. At the risk of labouring the point, it must be emphasized that these separate probabilities are only to be realized over a long series of trials.

Example 16.

A purse contains two half-crowns and three shillings; a second purse one half-crown and four shillings. A coin is drawn from the first and placed in the second, and then a coin is drawn from the second and placed in the first. Assuming that the chance of drawing a half-crown is twice that of drawing a shilling, find the most probable value of the coins in the first purse after the second operation.

At the end of the second operation the first purse may contain:

(i)z half-crowns and 3 shillings; or
(ii)3„ ²
(iii)r half-crown „ 4

The respective chances are found thus :

(i) In order to achieve this result, a coin of the same value must be taken from the second purse at the second draw as was placed in this purse as the result of the first draw.Since the chance of drawing a half-crown is twice that of drawing a shilling, the chance that a half-crown is drawn from the first purse originally is 4. The second purse will then contain 2 half-crowns and 4 shillings, and the chance of drawing a half-crown from this purse is . The total chance that a half-crown is drawn on both occasions is therefore -.

Similarly, the chance that a shilling is drawn both times is

.5.

The total chance that the coins in the first purse have the same value at the end of the two operations is therefore

4 4+ 3 5_ 2 9

r b—

By proceeding similarly, it can easily be shown that the respective chances under headings (ii) and (iii)

are,

and 4-

330 PROBABILITY
The greatest of these values is
The most probable value of the purse after the second draw is the value associated with this fraction, namely 8 shillings.
Example 17.

Cards are dealt one by one from an ordinary pack (without replaceuntil two aces have appeared. Find (i) how many cards (on the average) will be turned up, (ii) the most probable number of cards to be turned up.

(i) Since two aces cannot appear until the second trial at the earliest, we have
Chance of success at the second trial = -₅-⁴,. A.
third fourth

= ₂48 _{0 ,}3 er •^F481 =~•

48 47 4    •3
3    s i ~o s
and generally
Chance of success at the xth trial = ('C — 1) (52 — x) (5¹ - x) 4.3 52.51.50.49
The average number of cards to be turned up will be the sum of the series whose xth term is
x l(x — 1) (5² — x) (5¹ — x) 52.51 ^{4 .}3^0.49
for all values of x from 2 to 50 (since there must be two aces at least in the first 5o cards).
Summing this by ordinary algebraic or finite difference methods, it is found that the average number of cards to be turned up is 21^.2.
(ii) The most probable number to be turned up will be the value of x which gives
(x— ¹)(5^2—x)(5¹ ^—x)4^.3
52.51.50.49
its greatest value.
The value for x = r > the value for x = r — 1 so long as
(^r—1)(5^2—r)(51—r)>(r—2)(53—r) (5^2—r), i.e. so long as    — r² + 52r — 51 > — r² + 55r — 1o6,
i.e. so long as    55 > 3^r,
i.e. r is not greater than 18 (since r must be integral).
The most probable number of cards to be turned up is therefore 18.

EXPECTATION AND PROBABLE VALUE

11. Expectation and probable value.

The probability of happening of an event may be combined with a measure of the quantity depending on the event, thus producing an expectation. In other words, expectation is the product of an expected gain in actual value and the mathematical probability of obtaining such a gain. No special difficulty arises in this type of question, and the methods outlined above may equally well be applied to the solution of problems involving expectation. All that is necessary is to combine the respective chances with the gains in value should the events occur. Thus, if a player throwing

an ordinary die is to receive £2_n, where n is the number of throws that he takes to throw the first 6, his expectation would be
I I 5 I ^I5 5 I ^Ir5 r I I
6^£2⁺6 6^£2²⁺66^.6^£2^3+...+\6) •6£₂7+1+^..
the sum of which is £?-.

Probable value is the same as expectation, except that it refers to the things in question and not to the person. For example, in Example 16, the probable value of the coins in the purse at the end of the second operation (i.e. the expectation of the person drawing the coins) is

x 8 shillings + x _9zshillings + x 62 shillings = 74? shillings.
In other words, it would be worth while to give about 7s. 9d. for the purse after the second draw.
The following example is instructive :
Example 18.

A table is divided into six squares numbered I to 6. A player places a coin on a certain square. Three dice are thrown. If the number thus backed appears once, twice or thrice, the player receives back his own coin together with one, two or three others respectively of the same value. In any other event he loses his stake. Does the advantage in the long run lie with the player or the ^"banker^"?

(a) The chance that no die shows the number backed = (8)³.

(b)	one die	= 3 (s)z 1
(c)	two dice „	= 34^.G-)²
*(d)*	three dice	= (0^3.

332 PROBABILITY The net expectation of the player
=(b)+2(c)+3(d)—(a)=—A7z-, and that of the banker
=—(b)—2(c)—3(d)±(a)= ?l_g,
so that the advantage lies with the banker in the long run.
For a stake of a shilling it is easily seen that this advantage is just under a penny.

Here it might be contended that as the player is to receive back his own coin in addition to the prize, his expectation should be based on respective receipts of two, three or four units instead of one, two or three units, as appears in the above solution. His expectation on this basis might be argued as being 2 (b) + 3 (c) + 4 (d) — (a) or ,, which would show a substantial advantage to the player. Further consideration would show, however, that if the return of the stake is treated as a profit, then the laying of the stake must be treated as a payment to the banker for the privilege of playing and that it is, accordingly, definitely paid away whether the player wins or loses. His expectation then be-comes 2 (b) + 3 (c) + 4 (d) — 1 = — 11,5 as before.

12. In connection with mathematical expectation it is interesting to note the celebrated St Petersburg Problem, which has been a fruitful source of discussion for nearly two hundred years. Briefly stated, the problem is this : A coin is tossed until head turns up. If head turns up first A is to pay B one unit; if head does not turn up till the second throw B is to receive two units ; if not until the third throw four units, and so on. How much must B pay A before the game, in order that the game may be considered fair? That is, what is B's expectation?

The theoretical solution is as follows :
At the first trial B's expectation is _zx t    =
second    „    (1)² x 2    = 2^. ..................
15    nth    \].\ 1 n x 2n—1 = 1-.

B's expectation is therefore z + ? + ? + ... to infinity, and as this is a divergent series, it appears that B could afford to pay an infinitely large sum before the game starts for his expectation.

Many explanations of this result have been given by eminent mathematicians, notably by d'Alembert, Bernouilli and de Morgan. An inter-

THE METHOD OF INDUCTION 333

esting account of the problem with an alternative solution depending upon the amount of money that B possesses at the outset is given by Whitworth in his Choice and Chance. This solution depends on a somewhat arbitrary assumption, and a more practical limitation of B^'s expectation arises from the fact that A^'s resources, however great, must be limited. Poincare (Calcul des Probabilites, p. 42) shows that if A^'s total assets are 2', B^'s expectation is I + ip. For example, if A possesses 2³⁰—which is more than a thousand millions—B^'s expectation is reduced from infinity to i6, and as Poincare drily remarks, this is a considerable reduction!

13. The method of induction.

There is a certain type of problem in probability for which it is not possible to obtain a solution by the direct methods outlined in the preceding paragraphs. In these problems it is necessary to find a relation connecting the chance at any stage with that at sucstages and then to calculate the required probability by adopting an inductive process. The difficulty in these questions is to ascertain the fundamental relation : when this has been established the problem can be solved by the application of algebraic methods.
Since we have to obtain a relation connecting the probability of an event at the nth stage with those at succeeding stages, it is often of advantage to investigate the relation for simple numerical values of n and then to deduce the general result. The following is an excellent example of this method of solution.

Example 19.

Five green balls and sixteen red balls are placed in a bag. A ball is drawn at random n times in succession and replaced after each drawing. Find the chance that no two successive drawings shall have given green balls.

To satisfy the required conditions the events must have taken place as follows :
At the first draw either a red or a green ball must be drawn: R

At the second draw one of the following draws must have taken place: R R R G
G R
i.e. first drawing associated with R; or R G.

334
At the third draw:

PROBABILITY

R R R    R
    G R¹R    G R G R
i.e. second drawing associated with R; or first drawing associated with R G.
At the fourth draw:
    RRRR    RRRG
    GRRR    GRRG
    RGR,R    RGjRG
R R G^!R
G R G^'R
i.e. third drawing associated with R; or second drawing associated with R G.

It is evident therefore that if we have attained success at the (n — I)th stage, i.e. if no two successive green balls have been drawn, we shall attain success at the nth stage provided that, either

(i) the nth draw gives a red ball;
or (ii) if the nth draw gives a green ball the drawing at the (n — I)th stage gave a red ball.
The chance of drawing a red ball = and the chance of drawing a green ball = 2₁.
If therefore u_nbe the required chance after n drawings, we shall have
— 1 fi 7 f_i ^u,z — _7¹¹n_1 + ^un_2
This is a relation connecting three successive coefficients of the series

uo + u₁x + u₂x² + ... + u_n-2xⁿ-² + un_1xn ¹+ u_nxⁿ + ... .

The series is therefore a recurring series with scale of relation
    I — Iex —    ia_xs
Also u_o = 1 and u₁ = 1, since the conditions are satisfied if no ball or one ball is drawn.
Proceeding in the usual way, we find that the generating function is
    1[_D + x (11₁ — 16u₀/21)    _orI + 5x/21
(I + 4421) (I — 20x/21)    (I + 4X/21) (I — 20x/21)
25    I    I    I
24 I — 20X/21 24 I + 4x/21
— 25 \^I
loxl ¹1    4xl -1
24    21/    24\¹⁺221/
R G R G

ILLUSTRATIVE EXAMPLES    335
The required chance, u_n, is the coefficient of xⁿ in these two expanThis is easily seen to be
[²5 ( i)n — (— 1)n (MI^.

Note. As an alternative method of obtaining the coefficients of (? ⁰⁰)ⁿand (— i41)n we may say that u_n = a (Mⁿ + b (— ₂⁴,)ⁿ. a and b can then be found from the conditions u_o = I and u₁ = 1.

The following examples are illustrative of the same method of attack.
Example 20.

A player tosses a coin and is to score one point for every head turned up and two for every tail. He is to play on until his score reaches or passes n. If p_nis his chance of attaining exactly n, show that

P. = (^pn—1 + ^pn—2)

and hence find the value of pn .
There are two ways of reaching n exactly, namely, by throwing

(i)a tail when the score is it — 2; or

The respective chances are

_{2pn_2}

and

4pn_1

• Since these are mutually exclusive, we have

(

n—1

n—2)•

The value

may be found in either of two ways:

(a)pn = (^pn—1 + ^pn—2),

+ 2

n—1

pn—1

1-pn—2

(I).

= 3 }

obviously

(

₂

)

P2 =

By repetition of

(I)

and use of the facts in

(2)

we find that

pn =

I —

+ 4 — + ... + (—

^n—2

+ (—

1)n_I -,

2n—2

₂

which simplifies easily to

I-I {

2 +

(1)n

₂

(b)We may treat 1 — Z^x— 2x2 as the scale of relation of a recurring series and proceed as in Example 29. If the series is

p2x

+ p3x

+ ... + pn_2x

^n—2

+ pn_I

^xn—1

+ pnx

ⁿ

+ ...,

⁼

2=4

^and

^=1.

The generating function

the series is

I—

ix —

^—

x)(I

+Zx)

Also and

336 PROBABILITY
By partial fractions this becomes

2 I 3(I—x) 3(I+2x)^.

Expanding by the binomial theorem, we obtain for the value of pn (the coefficient of xⁿ) the same result as is found by the first method.
Example 21.

A has Io counters and B has 5; their chances of winning a single game are in the ratio 2 : I. The loser in each game is to give a counter to his opponent. The game stops when one or the other has lost all his counters. Find A's chance of winning all B's counters.
Let u_n be A^'s chance of winning all B^'s counters when A has n counters. In the next game A must either win or lose a counter. His chances of these contingencies are _iSand * respectively. When he has lost the next game his chance of winning all B's counters is _{un_1}and when he has won the next game it is ^un+I

Hence ^un = — z
S^un+I + 3¹¹n-1•
It is required to find u_n from the above relation, given that u₁₅= (since A will then have won) and u_o = o (since he will have lost). The required relation for the recurring series is therefore
-x+3x². Since we may write this relation as

3 (I — ^x+i^x2)⁼ 3 (I — ix)(I — x),

we may put u_n = a (2)ⁿ + b, and then obtain a and b from the values of u_D and ^uls .

This is a particular example of the problem of "duration of play." The problem in its general form gives m counters to A and n counters to B, and states that A's chances of winning, drawing and losing a single game are p, q, r respectively, where p + q + r = I. The method of solution is precisely similar to that above.
The inductive method can be adapted to other types of question in probability. An excellent example will be found in J.I.A. vol. LvI, pp. 102-104, where a problem in direct probability is solved very simply by the inductive process.

14. We will conclude this chapter with some miscellaneous exincluding among them questions involving probabilities of living or dying. These problems require only a careful application of the ordinary methods, and no special comment is necessary.

ILLUSTRATIVE EXAMPLES 337
Example 22.

Three men were known to be alive five years ago when their ages were 31, 48, 69. Assuming that of 98 males born together one dies annually until there are no survivors, find the chances that

(1)all are alive now;

The chance that a man aged x dies in five years is clearly 98—x' The required probabilities are

(

)

,s:

⁵

9 i

"°+

two similar expressions;

2 ? "

+ two similar expressions.

As the above are the only possible events that can happen, the total of the chances is unity and the reader should verify this.

Example 23.

Three men, P, Q and R, are each of exact age 96. Find the chance that they will all die at different ages last birthday in the order P, Q, R given that

Exact age	Chance of dying beforenext exact age
9⁶	i
97
9⁸
99	^I

We have to find the chance of their dying in a given order. There will be no need therefore to take into consideration the different orders in which the men may die.

(i) P may die at age 96    Chance = z•
Q may die at age 97, i.e. may survive
a year and die in the following year    = (I — 2)
R may die at age 98 or later ; i.e. may
survive 2 years and die in the
following year    = (I — (1 — 3) 1. The total chance under this heading is therefore
n
L ^x(^I2) 3 x (I    ) (1    3) I — 3c•

8    22

338    PROBABILITY
    (ii) P may die at age 96    Chance =
_ a
^R„    99    ,,    = (1 — 2) (^I— *) (I — 4) I.
Total chance under heading (ii)
= ^x(^I(^I1) ^x(^I2) (^I3) (^I— I) = 34T^.

(iii) P may die at age 97Chance = (I — ) 3.

Q„9

⁸

—

(

^I—

„

(

—

)

(

—

)

(I —

Total chance under heading (iii)

= (I —

(

2) (

DI-

—

(

3) (I —

= 6 q~.

Required chance =+ ;;

z-llv =

Example 24.

The sum of two positive quantities is constant and equal to 2n. Find the chance that the product of the two quantities is not less than ₄their greatest product.

The product of two positive quantities whose sum is constant is

greatest when the quantities are equal. The greatest product is therefore

271

2 2

If the two quantities are

and

2n —

we must have

ⁿ²

x(2n—x),

i.e.

—

8nx

i.e.

(zx —

3n) (2x — n)

Therefore

must lie between

³ⁿ

and

ⁿ

The possible values of

range from o to

2n,

and the chance required is therefore

3n n)

2n 22f

Example

25.

What is the chance that a hand of 5 cards contains a pair of

like cards of different suits?

This example is an excellent illustration of the manner in which a simple probability question can be capable of more than one reading. If the problem is to find the chance that of five cards two are to be like cards and the other three unlike, a result is obtained which is quite

ILLUSTRATIVE EXAMPLES 339 different from the chance that there are to be two like cards, it being immaterial what the other cards are. Moreover, this second reading is
capable of two alternatives: there may be two like cards and two other like cards (e.g. two kings and two fours) in the five cards; or there may be three or more like cards—this is not expressly excluded by the question.
It will be instructive before setting down all the chances that are
possible to examine what would seem to be the obvious solution. Let N denote the number of ways of selecting five cards from 52. Then 1\ 52.51.50.49.48
5•
The probability of drawing exactly two like cards of given denomina(aces, say) will be
4^.3 _X4^8.47^.4⁶
2! 3! 2162
or —
'
54¹45

and since there are 13 different cards in each suit it might be thought that to multiply the above fraction by 13 would give the result. This answer is, however, incorrect, in that there is "overlapping," for the remaining three cards out of the five may be a pair of like cards which have already been counted.

The possible arrangements, with their respective chances, are
Chance
N
Arrangement
48 x 13
13x⁴'³1^z X12X _-4^.3 -
(b) 3 like cards and 2 like cards 3 2
N
~3 X 4.3.2 _{X 42 X}12.11
3! 2!

(a) 4 like cards and 1 different
(c) 3 like cards and 2 different

(e) 2 like cards and ₃different - . 1760
N
or 4165
The reasoning to obtain these chances is straightforward, and it will be necessary to examine one only: the others are on the same lines.
* 44 because 8 cards are unallowable, being 4 of each of the previous numbers chosen. This avoids duplication with (b).

34⁰ PROBABILITY

Consider the arrangement in which there are to be exactly two like cards and consequently three different ones. Having settled on the denomination of the like cards—which can be done in 13 ways—there remain 12 denominations from which to choose 3. This can be done in I2. u . Io ways. The two like cards can be selected from the four

3
(one of each suit) in {'' ways. This does not quite complete the selec-2
tion, for the three different cards may be selected from any of the four suits, and this can be done in 4 ^x4 ^X4 ways.

Returning now to the analysis above. If the problem is confined to the chance where exactly two like cards are to be chosen, the result will be (e); if there are not to be more than two like cards, two sets of two like cards being permitted, the chance will be (d) + (e) ; if there are no restrictions, and we are given that two like cards at least are to be among the five, we shall require the total of the chances (a) to (e).
These results are respectively -} _ ; , and The above method of solution is designed to show, in full, the several probabilities of the various arrangements that may occur. If, however, it is known that there are no restrictions other than that there shall be
at least two like cards, a simpler method of solving this problem obviously presents itself. It is sufficient to calculate the probability that there shall not be any like cards among the five; this probability is evidently the complement of that required. The probability that there are no like cards is

13.12.11.10.9
5¹
x 45 or N
the complement of which agrees with the third answer given above. Example 26.

Ten clubs compete annually for a cup which is to become the absolute property of the club which wins it for three years in succession. Assuming that all the clubs are of equal skill, find the chance that last year^'s winners, not having won the previous year, will ultimately win the cup.

Let ^x be A^'s chance of winning outright, A having won the previous
I0
year. Each of the others has a chance equal to
1 — - 10
9    '
since the clubs are of equal skill.
2112
4163

EXAMPLES    34¹
Now (a) A's chance of winning next year and the year after _I/ I ²— ~Io⁾
(b) A^'s chance of winning, losing and then having the same chance as the other eight losing clubs
x
I —
I    q    t0
lo ' to '    9
(c) A^'s chance of losing and then having the same chance as
the other eight losing clubs
x
I - -
q    to
_
Io'    9
A's total chance = (a) + (b) + (c), and this we know to be Solving the resulting equation, we find that A^'s chance is -°.

Note. The above solution assumes that the total probability of winfor all the clubs, is unity, i.e. that the cup must eventually be won. It is easily seen that the cup must be won outright. If the chance that any particular club fails to win the cup outright after m trials is 1/n where n is greater than 1, then if there be an infinitely large number of sets of trials, the chance that the cup is never won outright by any

₁km
particular club will not be greater than Lt -) which is obviously
n,

zero. In other words, if the contests be continued for a sufficient length of time, the chance that the cup is not won outright is zero, i.e. the cup must be won.

EXAMPLES 18

Explain how the probability of a compound event, consisting of two constituent simple events, is obtained. Illustrate your answer by examples.

The chance of one event happening is the square of the chance of a second event, but the odds against the first are the cube of the odds against the second. Find the chance of each.

If three squares are chosen at random on a chess board, show that the chance that they should be in a diagonal line is

A man has three current English coins. Find the chance that he can give change for half-a-crown.

x to

34² PROBABILITY

A can hit a target four times in 5 shots; B three times in 4 shots; C twice in 3 shots. They fire a volley; what is the probability that two shots at least hit?

stand in a ring with ten other persons. If the arrangement of the twelve persons is at random, find the chance that there are exactly three persons between

and

The first twelve letters of the alphabet are written down at random. What is the probability that there are four letters between the A and the B?

Find the chance of drawing two white balls in the first two draws from a bag containing five red and seven white balls, balls drawn not being replaced.

q. An experiment succeeds twice as often as it fails. Find the chance that in the next six trials there will be at least four successes.io. If an experiment is equally likely to succeed or fail, find the chance that it will succeed exactly

times in

trials.

ii.

Find the chance of throwing ten with four dice.

If a die whose faces are numbered from i to 6 is thrown four times, what is the probability that the sum of the four throws is 14?

A five-figure number consisting of the digits o, I, 2, 3, 4 (no repetitions) is chosen at random. What is the chance that it is divisible by 4?

Out of a bag containing thirteen balls, six are drawn and replaced, and then seven are drawn. Find the chance that at least three balls were common to the two drawings.

If a die is thrown five times what is the probability that a six appears on at least two consecutive occasions?

What is the chance that a person with two dice will throw aces exactly four times in six trials?

There are

candidates taking an examination paper of

quesof equal difficulty; assuming that a candidate answers a question correctly or not at all, either being equally likely:

In how many different ways may a paper be answered?

a particular question is solved by not more than one candi

date?

EXAMPLES 343

Six cards are chosen at random from a pack of 52. Find the chance that three will be black and three red.

A card is chosen at random from each of six packs of cards. Find the chance that three will be black and three red.

The 26 letters of the alphabet are placed in a bag. A and B alternately draw a letter from the bag, the letters drawn not being replaced. The winner is the one who draws most vowels. A starts and draws a vowel with his first draw. What is his chance of winning?

A book contains i000 pages. A page is chosen at random. What is the chance that the sum of the digits of the number on the page is nine?

A bag contains three tickets marked with the numbers oo, 01, to, and two tickets each marked with the number i i. A ticket is drawn at random eight times, being replaced each time. Find the probability that the sum of the numbers on the tickets thus drawn is 33^.

If x be one of the first hundred numbers chosen at random, find

the probability that x +

too

is greater than

50.

In a lottery there are i000 tickets numbered i to i000. Three tickets are drawn. Find the chance that

(I)

the three tickets bear consecutive numbers;

(2)

two of the three bear consecutive numbers.

If m odd integers and n even integers be written down at random, prove that the chance that no two odd numbers are adjacent to one another is

n!(n+I)!

^--

n) ! (n - m +

t)!

being not greater than

The sum of two whole numbers is too; find the chance that their product is greater than moo.

There are ten tickets, five of which are numbered 1, 2, 3, 4, 5 and the other five are blank. What is the probability of drawing a total of to in three trials, one ticket being drawn and replaced at each trial?

If two of the first hundred numbers are chosen at random, what is the probability that their difference is greater than to?

A and B have equal chances of winning a single game; A wants two games and B wants three games to win a match. Find the chance that A will win the match.

344 PROBABILITY

A and B play at a game which cannot be drawn. On the average A wins three games out of five. Show that it is more than 2 to I that A would win at least three games out of the first five.

A, B, C throw in order, each using three dice. Prove that A^'s chance of throwing io first is (_T'',)² and find C's chance.

A and B play for a prize. A is to throw a die first and is to win if he throws 6. If he fails, B is to throw and win if he throws 6 or 5. If he fails, A is to throw again and win if he throws 6 or 5 or 4, and so on. Find each player^'s chance of winning.

and

play for a stake which is to be won by him who makes the highest score in four throws of a die. After 2 throws A has scored

and

9. What is

s chance of winning?

34• A and

play a set of games, to be won by the player who first wins four games, with the condition that if they each win three they are to play the best of three to decide the set. A

s chance of winning a single game is to

s as 2 to

Find their respective chances of winning the set.

A, B and C draw in succession from a bag containing four white and eight black balls until a white ball is drawn. What is the probability that the white ball is drawn by C? Is his chance improved if each ball is replaced after drawing?

and C, play a series of games and the winner of each game scores one point. Each of the three keeps a separate score and the winner of the set is the one who first scores 4.

wins the first,

the second and

the third game. What is then C

s chance of winning the set?A and

play a match of five games. A

s chances of winning,

drawing and losing any game are in proportion to 3, 2 and i respectively.

Two points are scored for a win and one for a draw. What is the chance that the match is drawn?

and

play a match, the winner being the one who first wins

two games in succession, no games being drawn. Their respective chances

of winning any particular game are

p : q.

Find

(I)

A's initial chance of winning;

(2) A's chance of winning after having won the first game.

Three players, A, B, C, play under the following conditions. In each turn the chance of success is the same for each of two contestants. A and B play together for the first turn, the winner plays with C, and

EXAMPLES 345
if he win again he wins the game; if not, C plays with the third man and so on until one man has won two turns in succession. Find each man's chance of winning the game.

40.
The winner of a game is the one who first scores 4 points, but if both players score 3 points the game continues until one player has scored 2 points more than the other. A and B play; find A's chance of winning when the score is z—o in B's favour, being given that A is twice as skilful a player as B.

B, C, D

each throw two dice for a prize. The highest throw wins, but if equal highest throws occur (thrown by two or more players) the players with these throws continue. A throws to; find his chance of winning.

and

cut a pack of cards, the player who wins the cut six times to be the winner. A, having won four times to

B's

once, cuts a five. Find the chance that A will be the winner.

43• A and B play a match consisting of a maximum of nine games. The chances that any game is won by A, won by B or drawn are equal. A win counts one point and a draw half a point (to each). The match ends when one of the players has a sufficient lead to leave him with an excess of points over his opponent even if the latter were to win all the remaining games. What is the chance that the match ends with the 7th game and not before?44^. Two persons throw an ordinary die alternately, and the first who throws 6 is to receive eleven shillings; find their expectations.X 45. There are eleven tickets in a bag numbered t, 2, 3, ... I I. A man draws two tickets together at random and is to receive a number of shillings equal to the product of the numbers he draws; find the value of his expectation.

46.

A bag contains eighteen exactly similar counters. Ten counters are each of value a, four are each of value b and the remaining four are each of value 2b. A man draws two counters at random. It is equally likely that any counter will be drawn. Find the ratio of a to b, if the value of the man^'s expectation is to be 4a ¹3.

47.
Each of two bags contains m sovereigns and a shillings. If a man draws a coin out of each bag, is he more or less likely to draw two sovereigns than if all the coins were in one bag and he drew two coins?

34⁶ PROBABILITY
48. Purse A contains six shillings and two sovereigns, purse B seven shillings and one sovereign. Seven coins are transferred from A to B and then seven coins are transferred from B to A. Which purse is now likely to be the more valuable?
49. A bag contains thirteen counters marked with the squares of the first thirteen natural numbers respectively.

A man draws a counter and is to receive the number of shillings equivalent to the number on the counter. Find his expectation.

50. A purse contains five half-crowns and four shillings. A pays 5s. 3d. for the right to receive the value of three coins drawn at random. Criticize his bargain, and find the chance that, after two attempts, the second on the same terms as the first, he will be a winner.51. A bag contains

white balls and two red balls. The balls are drawn from the bag one at a time without being replaced until a red ball is drawn. If

2, 3, ... white balls are drawn, A is to receive

, 3

, ... shillings respectively. Find his expectation.52. A bag contains twenty shillings and three sovereigns. Coins are drawn in succession, one at a time, without being replaced, until two sovereigns have been drawn. What is the probable number of shillings left in the bag?53

A bag contains a coin of value M, and a number of other coins whose aggregate value is m. A person draws one at a time till he draws the coin of value M. Assuming it is equally likely that any particular coin is drawn, find the value of his expectation.54• A bag contains twenty white balls numbered i to 20 and ten unnumbered red balls. A ball is drawn at random and replaced, six times. Find the probability that

(I)

at least three white balls are drawn;

(2)

three white and three red balls are drawn;(3) three red balls and Nos.

2, 3 of the white balls are drawn.

What is the most probable number of white balls drawn, and what is the probability that this number is drawn?

A bag contains ten counters, numbered i to ro. One counter is drawn and replaced and this operation is repeated until four different numbers have appeared. Calculate the probability that success will be attained with the sixth draw.

EXAMPLES 347
56. If n is the product of any 69 integers taken at random, find, to the first significant place of decimals, the value of the probability that n is not a multiple of 5, given that log_lo 2 = •3010300.
57. A bag contains counters marked with the digits 2, 4, 6, 8 and the number of times each digit occurs is equal to the value of the digit. Counters are drawn one at a time, each counter being replaced when drawn. What is the chance
(r) that the digit 2 is drawn before the digit 8;

(2)
that the sum of the first three digits is 16;

(3) that the first five counters drawn contain one of those marked 4or6?58. A and B have each eight pennies. Each tosses his set of pennies. Find to three places of decimals the chance that the number of heads obtained by A exceeds the number obtained by B by at least three.59. At a certain age 99 per cent. of the persons alive at the beginning of the year will live to the end of the year. Find expressions for the probabilities that out of four persons of that age there will die within the year

(1)exactly 2;

6o. If the probability that exactly three lives out of six all aged x survive n years is ^.o8r92, find the probability that at least three survive n years.

The probability that exactly one life out of three lives aged 20, 35 and 50, will survive 15 years is •092; the probability that all will die within 15 years is ^.00h. If the probability that the life aged 20 will survive 15 years is '9, find the probability that

(r) he will survive

years;(2) he will survive 45 years.

Three men, A, B, C, are each aged 30. Given that the probability that a man aged 30 will survive 5 years is •974 and the probability that he will survive to years is •940, find the chance that between the end of the 5th year and the end of the loth year from now

(r) one at least will die;

(2) all will die,

dying first and

second.

Find also the chance that

and

will die within this period and C will survive the roth year.

348 PROBABILITY

63. Given that the probability that of three lives aged x one, and one only, will survive n years is 27 times the probability that all will die within n years, find the probability that

(a)at least two will survive n years ;

years.

64. Given the following table, find the probability that of four persons aged 65 at least one will die between ages 75 and 85 and at least one after age 85 :

AgeProbability of surviving io years

65One-half75One-fifth

65. If m things are distributed amongst a men and b women, show that the chance that the number of things received by the group of men is odd is equal to

(b + a)ⁿi — (b — a)^m

66. The sum of two positive quantities is constant and equal to 2n. What is the chance that their product is less than

67. A purse contains four half-crowns, three pennies and two shillings. Four coins are drawn at random. How many different sums can these amount to, and what is the most probable sum? (Assume that any one coin is as likely to be drawn as any other.)

68. How many times must a man be allowed to toss a penny so that the odds may be loo to i that he gets at least one head?

69. A coin is tossed m + n times (m > n). Prove that the chance of at least in consecutive heads appearing is (2 +~ Find also the chance of a run of exactly m consecutive heads.

70. If ten different things be distributed among three persons, show that the chance of a particular person having more than five of them

71. If

be the chance that an odd number of aces turn up when

ordinary dice are thrown, show that

1 —

(3)

ⁿ

72. A pack of cards has been dealt in the usual way to four players. One player has just one ace; prove that the chance that his partner has the other three aces is

73. A bag contains a certain number of balls some of which are white. I am to get a shilling for every ball so long as I continue to draw white

EXAMPLES 349
only, the balls drawn not being replaced. An additional ball not being white is introduced and I claim as compensation to be allowed to re-place every white ball that I draw. Is this fair?
74. There are three sets of cards, red, yellow and blue. Each set contains ten cards, numbered i to io. Three cards are drawn at random. Find the chance that the sum of the numbers on them equals 15 :
(r) if the cards are all to be drawn from the red set;

(2)
if one card is to be drawn from each set;

(3) if the cards are to be drawn from the three sets mixed indis75. From a bag containing ten red, ten white and ten blue balls one is to be drawn at random and replaced. The operation is to be repeated ten times. Find the chance that at least one ball of each colour will be drawn.76. There are ten counters in a bag marked with consecutive numbers. Two counters are drawn from the bag. If the sum of the numbers drawn is odd, a man is to receive that number of shillings; if it is even, he is to pay that number of shillings. Find the man^'s expectation

(I) if the counters are marked from o to 9;

(2)if they are marked from r to ro;

II.

77. Out of 3n consecutive integers three are selected at random. Find the chance that their sum is divisible by 3.78. In a book of values of a certain function there is one error on the average in every m values. Find the number of times, r, a value must be turned up at random in order that you may have an even chance of turning up an erroneous value. Show that when m becomes large the ratio rⁱm tends to a fixed quantity and find this quantity.79. A, B, C and five other football teams enter for a competition. The teams are of equal skill and are drawn by lot in pairs before each round, the winners of the previous round entering the next round. Find the chance that in the course of the competition A will beat B, having first beaten C.80. A street consists of 24 houses numbered r to 24, odd numbers on one side, even on the other. Three houses are vacant. Assuming that the houses are all equally likely to be vacant, find the chances:

(1)

that the three houses are next to each other;

(3) that if they are all on the same side the sum of their numbers equals

350 PROBABILITY
Si. Find the chance that the sum of the numbers on three cards drawn at random from an ordinary pack of 52 cards amounts to 21, all the court cards counting as to. How will the result be altered if an ace can count as t or I1?
82. If a coin be tossed 15 times, what is the probability of getting heads exactly as many times in the first lo throws as in the last 5?
83. A bag contains eight counters numbered t to 8. Four are drawn at random. Find the chances that
(t) the sum of the numbers on the four counters amounts to at least 17;

(2)
the counters numbered 2 and 3 are among the four;

(3) the four counters contain at least two of the three numbers 3, 5, 7^.84. If 6n tickets numbered o, I, 2, ... 6n — t are placed in a bag and 3 are drawn out, find the chance that the sum of the numbers on them is equal to 6n.85. From a bag containing nine red and nine blue balls nine are drawn at random, the balls being replaced. Show that the probability that four balls of each colour will be included is a little less than86. A looks at a clock at some time between 2 and 5 p.m., all times within the limits being equally likely. He looks again when it strikes the next quarter hour. What is the chance that in the meantime the minute has overtaken the hour hand?87. Four suits of cards, each suit consisting of 13 cards numbered from I to 13, are dealt to four persons. Find the chance that each person^'s cards contain all the numbers from t to 13.88. There are four sets of calculations on one sheet which have to be made, then checked and finally scrutinized. A and B can calculate only, C and D can calculate or check, and E and F can scrutinize only. No person may check a calculation he has made and all work must be signed. Find the chance that when the sheet is finally completed each name of the above six appears exactly twice.89. From an ordinary pack of cards a card is drawn and then six other cards at random. Find the chance that the card first drawn is the highest of its suit amongst all the cards drawn.90. A coin is tossed until both head and tail have appeared twice. (I) On the average how many times will the coin have to be tossed?

(2)

What is the most likely of throws?

(3) How many throws must a man be allowed if the odds in favour of success are to be 7 :

EXAMPLES 351

A bag contains thirteen balls of which four are white and nine black. If a ball be drawn r times successively and replaced after each drawing, show that the chance that no two successive drawings shall have given white balls is

16.12r—(-3)"

15.13

The reserved seats in a certain section of a concert-hall are numbered consecutively from I to Ioo. A man sends for five consecutive tickets for one concert and for eight consecutive tickets for another. Find the chance that there will be no number common to both sets of tickets.

93, In a cup draw there were four Southern and four Northern teams, the names being drawn one by one from a bag. Find the folprobabilities:

(I)

that each one of the four successive pairs consisted of one Southern and one Northern team;

(2)

that each one of the four successive pairs consisted of two Southern or two Northern teams;

(3) that the first drawn in at least three of the pairs was a Southern team.

94. A number consists of seven digits whose sum is 59. Find the chance that it is divisible by

I1.

95• Two dice are thrown and one of the planers will win (a) if the sum be

₇

or (b) if the sum be

_4,

5, 6, 8, 9 or

and the same sum reappears before 7. Find the player's chance of success.

A and B cast alternately with two dice. It is agreed that, on each failure to win, the prize money is to be reduced by 3 per cent. of its value at the previous attempt. A wins if he throws 6 before B throws 7, and B wins if he throws 7 before A throws 6. A starts first. Compare the values of the respective chances of A and B.

A bag contains n counters marked

1, 2,

3, ...

If two counters are drawn show that the chance that the difference of the counters exceeds

(less than

n —

(n—m)(n—m—i)

n(n —

Deduce from this result the chance that if two points are taken at random on a line the length between them exceeds half the length of the line.

352 PROBABILITY

A bag contains two white balls and one black ball. A drawing of two balls is made. If either is black, the two are replaced and another black ball is added. A second drawing of two balls is then made, and again if either is black, the two are replaced and another black ball is added and so on.

What is the chance that if the drawings are continued indefinitely two white balls will never be drawn together?

A^'s chance of scoring any point is of B^'s. A engages to score 14 in excess of B before B shall have scored 3 in excess of A. Show that A's chance of winning the match is equal to

—

(-t)3

I - (1)17 .

Too. A and his wife engage in a "mixed doubles" tennis tournament in which each pair of players consists of one member of each sex. There are fourteen other persons, seven of each sex, also entered for the tournament and players are drawn by lot before each round in such a way that any person of one sex may be the partner of any person of the other sex. Only the winners in any one round enter the next round. Assuming that all the players of each sex are equal in skill, find the probability that in the final round A and his wife play together as partners.

Io1.

All that is known about a quadratic equation is that the coeffiare all different and are positive integers less than lo. Find the chance that the roots of the equation are real, all the integral values of the coefficients satisfying the above conditions being equally likely, and zero values of the coefficients being excluded.

102.

A pack of cards is dealt in the usual way to four players of whom

two and two are partners and the dealer turns up his last card. Denoting by the term " honours" the ace, king, queen, knave of the suit to which the turned-up card belongs, find the chance that each pair of partners shall have two honours.

103.

A number taken at random is squared. Find the chances that the following are even numbers :

(I)

the digit in the units place of the result;

the digit in the tens place of the result ;

104.

Before commencing a game of cards four players cut for partners,

i.e. the two highest play together and the two lowest together. All suits being of equal value, what is the chance that they will have to cut again?

EXAMPLES 353

1o5. A and B play a series of games to be won by the player who first wins two consecutive games. A^'s chances of winning, losing or drawing any particular game are + and _arespectively. Find B's chance of winning the match (a) at the outset, (b) when he has just won one game, and (c) when he has just lost one game.

io6. A and B have equal chances of winning a single game. A wants n games and B n + r games to win a match. Show that the odds on
A are i+ p to i— p where p= (zn)

—_n~_n22n•

107. Three posts are filled one after another by lot from amongst ten persons (A, B, C, D, and six others), the first by any one of the ten, the second by anyone except A and B, the third by anyone except C and D. What is the chance that A or B or both of them, and C or D or both of them, are chosen? No one can hold more than one post, and in drawing for the second and third posts the barred persons and any person previously chosen are excluded.

io8. With a hand of thirteen cards a player is known to hold one ace. What is the chance that he has at least one other ace?
If it is known that the ace he holds is the ace of hearts, what is the chance that he has at least one other ace?