Calculus and Probability for Actuarial Students

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CHAPTER XXI

PROBABILITY

In considering the subject of Probability a clear appreciation must be obtained of the distinction between its mathematical or theoretical treatment and its arithmetical or practical development which is the basis of actuarial science. The difference between these two points of view will become clear by examining a few simple illustrations.

Probability or chance is merely an expression of relative degrees of uncertainty in relation to an event in respect of which our knowledge is incomplete. In the ordinary phenomena of life probmay not have a numerical value although distinctions are drawn. Thus we should say, in the absence of more precise knowledge, that a man of 30 was as likely as not to die before another man of 30, would most probably outlive a man of 60 and would almost certainly outlive a man of 80. The mind has clearly formed definite opinions on these points but the respective probcannot be expressed in numerical form without further analysis.

To obtain a measure of probability some unit must be taken, and for reasons of convenience it has been the universal custom to take absolute certainty as having unit probability. Starting from this standard, certain probabilities can be found from general reasoning, whereas others, such as the chance of dying in a year, or of becoming the father of twins, must be deduced from the obof suitable statistics.

As an example of the former class, one may say that, on tossing a coin, it is certain that either "heads" or "tails^" will appear, and as these possibilities are equally likely, the chance that heads will appear is clearly i.

We cannot however assume from this that, if we toss a coin four

126 PROBABILITY
times, two heads and two tails will necessarily appear. Indeed, as will be shown later, the respective probabilities are
0 heads 4 tails
1 head 3 tails
2 heads 2 tails
3 heads 1 tail
4 heads 0 tails

What we understand is that, if the number of trials were sufficiently extended, the proportion of cases in which heads appeared would approximate more and more closely to .1. It is in this sense that the mathematical definition of probability is to be understood in connection with actuarial work.

The following Articles deal with the treatment of the subject from its mathematical aspect.

From the foregoing considerations it will be evident that if an event can happen in `a^' ways and fail in `b^' ways, and each of these

ways is equally likely, the probability of its happening is

and that of its failing is

For the sumof these two probabilities

gives the chance of its either happening or failing, which is necesunity.

Alternatively, we may say that

the probability of an event is equal to the ratio of the number of cases favourable to the event, to the total

number of cases.

It follows that if p is the probability that an event will happen, the probability of its not happening is 1 —p.

Another method of statement is to say that the odds are `a^' to `b' in favour of the event or `b' to `a^' against the event.

Example.

The chance of throwing a four at a single throw of a

die is -, for there is one favourable result (a 4) and five unfavourable

results (1, 2, 3, 5 or 6), all of which are equally likely.

The chance of

not

throwing a four isor, in other words, the
odds are 5 to 1 against the event.

PROBABILITY 127

6.The following proposition is also self-evident :

If an event can happen in more than one way (all ways being, however, mutually exclusive), the probability of its happening at all is the sum of the several probabilities of its happening in the several ways.Thus if the chance of scoring a " bull " be -, that of scoring an "inner " be that of scoring a " magpie " be 5 , and that of scoring an "outer" be , the total chance of hitting the target at all (all events being mutually exclusive) must be ₁¹₀- + 8 +—'~ + = 4_f',.

The solution of elementary questions in probability depends therefore upon general reasoning, but calculation is aided in some cases by the theorems of permutations and combinations.

Some elementary examples will now be given :

The odds in a given race against three horses are 11 to 4, 13 to 3, and 7 to 2 respectively. Find the chance that one of them will win the race, a dead-heat being assumed to be impossible.

The chance that the first horse should win is

is,

second

Tx,

third

Thus the chance that one of them should win is

4 .}

2 -

487

7~~

16-

A has three shares in a lottery where there are three prizes and six blanks. B has one share in another, where there is one prize and two blanks. Show that A has a better chance of winning a prize than B in the ratio of 16 to 7.

To be successful A may draw either 3, 2 or 1 prizes.He may draw 3 prizes in 1 way.He may draw 2 prizes and 1 blank in

(2)

(s)

= 18 ways.

He may draw 1 prize and 2 blanks in

(i)

(

₂

)

= 45

ways.

The total number of ways in which he can win at least one prize is therefore 1 + 18 + 45 = 64.

Now three tickets can be selected in

(

₃

)

= 84 ways.

Therefore A's chance of success is

L4 =

it.

B's

chance is clearly

Therefore A's chance is to

B's

chance in the ratio 16 : 7.

Alternatively

may draw all blanks in

(3)

= 20 ways. His

128 PROBABILITY
chance of non-success is therefore and his chance of success 1— s4 = as before.

(3)If four cards be drawn from a pack, what is the chance that there will be one from each suit ? Four cards can be selected from the pack in (4²) = 270725 ways.

Four cards can be selected so as to be one from each suit in 13° = 28561 ways.

Therefore the required chance is

-aTz13 = -

₃

Out of a bag containing 12 balls, 5 are drawn and replaced, and afterwards 6 are drawn. Find the chance that exactly 3 balls were common to the two drawings.

The total number of ways of making the second drawing is

(

₆

)

To comply with the conditions, it must contain 3 balls out of the first 5 chosen, and 3 balls out of the 7 left on the first choice.

The respective ways of making these selections are (

₃

)

and (

₃

)

The total number of selections favourable to the event is therefore

(53)

(

⁷

₃

)

= 350, and the required chance is

34 -

66'

Twelve persons take their places at a round table. What is the chance of two particular persons sitting together ?

Let the two persons be

and

Then, since we are only conwith the

relative

positions of the persons, we may regard A's place as fixed. There are then 11 other seats in all, 2 of which are adjacent to A. B's chance of occupying one of these is there-fore i

What is the chance of throwing more than 10 in a single throw with two dice ?

A score of more than 10 can be made by the following throws : 5 and 6, 6 and 5, 6 and 6.

The total possible number of combinations is 6 x 6 = 36 and, as 3 of these are favourable to the event, the required chance is

3 - 1

It should be noted that exactly 11 can be thrown in two ways, since (regarding the throws as consecutive) either a 6 or a 5 may appear first, but 12 can only be thrown in one way.

PROBABILITY 129
8. If the chance of an event is p and the measure of the quantity dependent on the event be x, the product px is called the expectation.
Thus we may have the expectation of a person in a lottery, or the expected value of a prize, or the expected number of deaths among a given number of persons, and so on.
Examples.

What is the expectation of a person who is to draw one envelope from a bag which contains one £1 note, two 10s. notes and three blank pieces of paper, each placed in an envelope of uniform size?

The chance of drawing the £1 note is

The chance of drawing a 10s. note is

The value of the expectation is therefore

x 20s. + .- x 10s. = 6s.

8d.

If the chance of dying in a year is , the expected number of deaths among 100 people is _-ax 100 = 5.

In the foregoing articles we have considered what are, in principle, single events. We have now to determine the appropriate formulas for combinations of two or more events.

In considering problems of this kind, a close watch must be made

to see if the events are dependent or independent.Thus, a bag contains 6 white and 4 black balls, and it is desired to estimate the combined chance of drawing at the first draw 3 white balls and at the second draw :3 black balls. If the balls are

replaced after the first draw, the second event is clearly independent

of the first. But, if they are not replaced, the drawing of 3 white balls at the first draw will obviously affect the chance of drawing 3 black balls at the second draw, and the two events will be dependent.

Dependent events are also called

contingent.

10. We are thus led to the following proposition:

The chance of two independent events happening is the product of

the

chances of their happening severally.

For if the first event can happen in

ways and fail in

b ways,

and the corresponding figures for the second event are

and

the

total possible combinations of events are

(a + b) (a

+ b

Of these

130    PROBABILITY
Both events may happen in    aa^' ways.
The first event may happen and the second event
fail in    ab^' ways.
The first event may fail and the second event
happen in    ba^' ways.
Both events may fail in        bb^' ways.
The chance of their both happening is therefore (a + b) aa'
(a' + b') or a a
b^. a' + b^, , i^.^e.the product of the respective chances of their happening.
If the respective chances are p and q,
the chance that both happen is    pq,
the chance that the first happens and the
second fails is    p (] — ),
the chance that the first fails and the
second happens is    (1 — p) q,
the chance that both fail is    (1 — p) (1 — q),
the chance that at least one happens is        1 — (1 — p) (1 — q)
^=p+^q—pal, the chance that one and only one happens
is    p(1 —q)+(1—p)q
=p+q—2p1.

The above results can be applied by a slight modification of reasoning to events which are not independent.

Thus if

be the chance of the event happening, and

be the chance of a second event happening

when the first has happened,

then the chance that both should happen is

pq.

By successive applications of the above reasoning we can arrive at a formula for any number of events.

Thus if p be the probability of an event A; and, when A has happened,

be the probability of another event B; and when A and B have happened, r be the probability of another event C; and so on for any number of events; the chance t,'tat all the events will happen

the product pqr....

It follows that if p be the chance that an event will happen in one trial, the chance that it will happen in each of a series of r trials is

and the chance that it will happen at least once in a series of

trials is 1 — (1 — p)''.

PROBABILITY 131

This result can be developed more generally in the following way:

If p be the probability of an event happening in one trial, what is the probability of its happening once, twice, three times, ... exactly in n trials?

Let

be the probability that the event does not happen, so that

p+q

= 1. Then if the event is to happen exactly

times, it must happen in each of a given combination of r of the trials and fail in each of the remaining

(n — r)

trials. The chance of this occurring is, as seen above,

^—

But the particular set of

trials can be

chosen in (

)

ways, each of which is equally likely. Therefore the total chance of its happening exactly

times is

(

)

^—r

or the

term containing

in the expansion of

+ q)".

Thus the successive terms of this expansion represent the probabilities of the event occurring respectively

n, n —

n — 2, ...

times in

trials.

The foregoing propositions can best be grasped by applying them to certain examples.

If four cards be drawn from a pack, what is the chance that there will be one from each suit?

Let one card be drawn, which may be of any suit. The chance that a second card is of a different suit is , for there are 51 cards remaining, 39 of which will be of different suits from the first.

Similarly the chance of drawing a third card of a different suit from

the first two is

and that of a fourth different card is

The combined chance is therefore

:,5

x 2

= 2o

This result should be compared with that obtained in Example 3 of 7, where the problem was treated as that of a simple probIt is clear that the same result must be obtained whether

the cards are treated as being drawn simultaneously or successively.

A man throws a six-faced die until he gets an ace; he is to receive £1 if he succeeds at the first throw, £i if he succeeds at the second throw, £A if he succeeds at the third throw, and so on; given that log, 6 =179176, find the value of his expectation.

The chance of succeeding at the first throw is The chance of succeeding at the second throw is compounded of the chance of failing at the first throw and succeeding at the second throw, etc.

132 PROBABILITY The value of the expectation is therefore ={-log, (1—a)}
=Slog, 6
= 358:1 = 7s. 2d.
(iii) Let it be assumed that the probabilities of dying within ten years after the ages specified are, on the average, as follows:

Age within next 10 years Probability of dying
30 40 50

i i
What is the chance

(a)that a person A now aged 30 should die between the ages of 50 and 60 ;

and

aged respectively 30 and 40, should be alive 10 years hence ;

that of two persons A and B, aged respectively 30 and 40, A should die between the ages of 40 and 50 and B should survive to the age of 60 ?

(a)

The required chance is compounded of the chances that A should survive successively to ages 40 and 50 and should then die within the next ten years.

The several chances are (1 —A), (1 --&) and

and the required

chance is the product of these three factors, namely

^{xgx =}

i2a•

(b)
The respective chances of surviving ten years are (1 — A-) and (1 - -). The chance that both A and B should survive that period is therefore

=14

(c)The chance that A should die between the ages of 40 and 50 is +1 x'~.

The chance that B should survive to the age of 60 is

The required chance that these two events should both happen is, therefore,

(+a ^{x x x}b)⁼A3•

PROBABILITY 13 3

(iv) The faces of a die are marked with the consecutive numbers 1, 2, ... 6. What is the chance that, after seven throws, the sum of the numbers exhibited equals 30 exactly ?

This example is important, as it illustrates a method which can frequently be employed.
The number of ways in which the seven numbers exhibited can total 30 is given by the coefficient of x³⁰in the expansion of (x+x2+...+xe)",
for this coefficient arises from the combination of the indices of x, taken together in such a way as to produce a total of 30.
Writing (x + x² + ... + x⁶)' as x" (1 — x⁸)' (1 — x)-" it is seen that we require the coefficient of ^x23in the expansion of

(1_x6)"(1—x)-'=(1—7x8+21x12—35x'5+...)(1—x)-".

The expression within the first bracket need not be expanded further, since no term higher than x²³is required. It remains to combine the given terms with appropriate terms taken from the expansion of the second bracket in such a way that the power of x given by the product of the two terms is 23.
Thus we get

(1) x 24 x ... x 29 (x231 = 475,020 _x23
1x...x6 I

(— 7x⁸) x (18 x ... x 6 x^l") = — 706,629 _x23

'l2x...x17
(21x¹²) x l x ... x ₆ x'¹) = 259,896 x
(— 35x") x (6 x ... x 11 x5) = — 16,170 x²³ 1x...x6 I
12,117 x"³
The number of ways in which a total of exactly :30 can be made is thus 12,117.

Now, on any of the seven throws, any of the six numbers may be exhibited. The total number of possible combinations of numis, therefore, 6'. The number of combinations giving a total

of 30 being 12,117, the required chance is 12,117 6'

134 PROBABILITY

15. The following miscellaneous examples are taken from the examination papers of the Institute ; they illustrate various devices which may be employed with advantage in the s lution of questions of this character.

(i)If n whole numbers be multiplied together, find the chance that the last digit of the product is a five. [1910, Paper I, Q. 7.]

The chance is compounded of the separate chances that none of

the final digits of the n numbers is even and that one at least is a five.

The required chance is therefore

10" —

(ii)
Four coins are tossed together and A is to receive £2 if exactly 2 heads turn up, and to pay £1 in any other event. Find
the probability that after four trials A is £1 out of pocket. [1913, Paper I, Q. 8.]

To satisfy the conditions A must win once and lose thrice.The chance that exactly 2 heads turn up equals the middle term

in the expansion of

+1)4 =

(2)

x (2)

⁴

Therefore the chance that he should win once and lose thrice equals the second term in the expansion of (g

)

⁴

which is

3 (4)' -

375

(iii)
If a coin be tossed 15 times, what is the probability of getting heads exactly as many times in the first 10 throws as in the last 5? [1915, Paper I, Q. 9.]

If n coins be tossed the chance that exactly r heads turn up equals the (r + 1)th term in the expansion of (1 +

2)71.

Now in the last 5 throws we may get 0, 1, ... 5 heads and we have to combine the chance of any of these with the chance of getting the same number of heads in the first 10 throws. Thus we have for the several chances of getting

0 heads in both sets of throws =	(I)⁵.(2)⁷⁰
1 head,,=	5(4)5.10(1)¹⁰
2 heads„,,=	10(2)⁵45(1)¹⁰
³=	10(I)⁵. 120(1)¹⁰
4	⁵(1)^5.210(2-)¹⁰
⁵	(1)^5.252(1)¹⁰

PROBABILITY 135

Therefore the total chance that the same number will turn up in both sets of throws is the sum of each of these distinct probabilities, 3003
i.e. gin .

(iv)
A and B throw for a certain stake, each throwing with one die; A^'s die is marked 2, 3, 4, 5, 6, 7, and B's 1, 2, 3, 4, 5, 6, and equal throws divide the stake; prove that A^'s expectation is ₄₄of the stake.

What will A

s expectation be if equal throws go for nothing? [1911, Paper I, Q. 8.]

Let e equal A

s expectation.

The chance of A's throwing any of the numbers marked on the die is the same for each number.

If A throws a 2, B must throw a 1 if A is to win the stake, or a 2 if A is to divide the stake. The chance that A throws a 2 is -; the chance that he then wins the stake is - and that he divides the stake is also i. His expectation if he throws a 2 is therefore s + - . 2 .If equal throws go for nothing, A^'s expectation after an equal throw clearly remains at e. In that case his expectation if he throws a2isi+e.

'We therefore have the following scheme:

A's throw I Chance	A's expectation if equal	A's expectation if equal
thereof	throws divide the stake	throws go for nothing
2		+- e*
3		A+'_Ae
4		ie
5		;+e
6		+e
7

The total expectation being the sum of the separate expectations, we have in the first case e = , being the sum of the figures in the
ird column of the above table, and in the second case e = + he.
Whence e=1.

(v)
A man tosses 20 pennies and removes all that fall head up; he then tosses the remainder and then removes all that fall head up, and so on. How many times ought he to be allowed to repeat this operation if he is to have an even chance of removing all the pennies before he has finished? [1906, Paper I, Q. 11.]

136 PROBABILITY

The problem is clearly the same if all the pennies are tossed each time; we then have to find the chance that all the pennies have turned up heads once at least.

If n be the required number of throws, the chance that any particular penny has turned up heads once at least is {1 — (')ⁿ}.

The chance that all the pennies have turned up heads once at least is therefore {1 — ()ⁿ}²⁰and by the terms of the question this must equal i.

Solving this equation we find (2)ⁿ = — (I)²⁰,

n = 4^.87.

(vi) 2ⁿ pla,rfe-of equal skill enter for a tournament; they are drawn in pairs, and the winners of each round are drawn again for the next. Find the probability that two given competitors will play against each other in the course of the tournament. If n = 5, show that the probability that a given player will either win or be beaten by the actual winner is A. [1912, Paper I, Q. 6.]

As regards the first part of the question, the total number of games that will be played is

2n-1+2n—a+...+1=2n—1.

Also two players can be selected in 2ⁿ (2ⁿ — 1) ₂ ways. The chance that two will meet is therefore the quotient of these two values,
1 i.e.

The second part of the question can be proved by an inductive process.
Let u_n be the required chance.

Then in the first round he may either win or lose. If he wins, he passes into the next round and his chance of winning or being beaten by the ultimate winner then becomes u_n_₁. If he loses, his opponent passes into the next round where the latter's chance of becoming the ultimate winner is increased to 1__i.

We thus have
1 1 1

'un = ~ "n-1 + . qn-1 .

PROBABILITY    137
1    1    1    1
Similarly    u„_, = ₄un-a + 4 .
1    1    1    1
^un—2 = g ^un—3 + 8 • 0 _tn_₃

etc.    etc.
1    1    1 1

2n—2 ufl = 2^{n—i ui}+ 2^n—' 2

As u, obviously equals 1, we have by addition

1 n-1 n+1

4Gn = 2n-i + 2" = 2n v

U5 =

It is instructive also to prove the first part of the question inductively.

Let u_n be the required chance. This chance is compounded of the two chances that they meet in the first round, or that, not having met in the first round, they both pass into the second round where the corresponding chance becomes un_i.

The chance that they meet in the first round is on      ; the chance
2ⁿ-2
that they do not meet is _2n-    . Therefore we have
1    1 2ⁿ-2 u_n= 2" — l + ^un—i• 4 • ₂n — ₁,
2^n—i— 1
or    (2ⁿ — 1) u_n=1 +    ₂^un-i•
1 2^n_2— 1
Similar/    (qn~    ^—i2 - 1) u_n_, = ₂+    4 ^un—2,
/,    1^/1 2^n_3 - 1
(271-2- ₄) ^un—2 = 4 +    8    ^une,
etc.    etc.
(2^a -1)    1    2²-1 2n—3 ^u3= 2n—3 + In—2 ^u2,
(2²-1    1    2-1 (2'2n--a ) ^us = _2n-2 + 2n_i u

138    PROBABILITY
By addition
(2ⁿ-1)u, =1+ ¹+4+...+₂1_₂+_2,1_i (since u_l obviously equals 1)
1

1 -
2ⁿ 2ⁿ—1

1 1
_

2
1

Whence un =
as before.
16. The application of the Integral Calculus to problems of mean value and probability is shown in Chapter XIX, % 5-7.