A Popular Course of Pure and Mixed Mathematics ...

In the same manner it may be shown that the number of parcels or permutations in four's is m47; therefore, in general, the number of parcels or permutations with m things, taken n at a time, is mi.

(2.) Corollary :-When n is equal to m, the number of permutations of m things, taken m at a time, is mm: 7=1m7; hence the per. mutations of two things in two's is 1x2=2; of three things in three's, is 1x2x3=6; and of four things in four's, is 1 X2 X3 X4= 24; and so on.

Theorem 2. (3.) The number of combinations, without repetitions, raised from a

base taken m things, taken n things at a time, is

m217

The number of permutions for m things in two's is melī; but the permutations of two things is 1.2=2; therefore two things has 12=2 more permutations than combinations, and consequently in the whole number malt of permutations there are 1.2 times more permutations than combinations; therefore the number of combinations of m things, taken two at a time, are

1211 In the same manner, as every parcel of three's admits of 1.2.32 1341=6 more permutations than combinations, but the permutations of me things in three's is m317; therefore the number of combinations of on things in three's is

978311
1311°

In general since the number of permutations of m things taken R at a time is manī, and the number of permutations of n things taken n at a time is 1"is, the number of combinations raised from a base of

(4.) The number of parcels, with n things in each, having repe.

titions raised from a base of r things, is

For the number of parcels with n things in each, having no repe

from a base of m things, is

Now m and m-ntl

are the first and last terms of the factorial mais, therefore

(m-n+1)

But the number of things in the base required to raise a class of combinations with n things in each parcel without repetitions, is greater by (n-1) than the number of things in the base required to raise the same number of parcels with the same number of things in each, having repetitions. Therefore, deducting n-1 from m, gives m-n+1 for the number of things in the base

(m-n+1)* required to raise

parcels with n things in each.

grli Now let r=m-n+1; then will be the number of parcels with n things in each, raised from a base of r things.

ON CHANCES.

(1.) If an event may take place in n different ways, and each of these be equally likely to happen, the probability that it will take place in a specified way is properly represented by certainty being represented by unity: or, which is the same thing, if the value of certainty be unity, the value of the expectation that the event will

is For, the sum of all the probabilities is certainty, or unity, because the event must take place in some one of the ways, and the proba

1 bilities are equal; therefore each of them is

(2.) Cor. If the value of certainty be a, the value of the expectation is But in the following articles we suppose the value of certainty to be unity.

(3.) If an event may happen in a ways, and fail in I ways, any of these being equally probable, the chance of its happening is

and the chance of its failing is a+b?

a +7 The chance of its happening niust, from the nature of the supposition, be to the chance of its failing, as a tov; therefore the chance of its happening, together with the chance of its failing :: 2 :a +b; and the event must either happen or fail ; consequently, the chance of its happening, together with the chance of its failing, is certainty; hence, the chance of its happening : certainty :: a: +b; and the chance of its happening=

at Also, since the chance of its happening together with the chance of its failing is certainty, which is represented by unity, 1

at6 b eft is the chance of its failing

. (4.) Er. 1. The probability of throwing an ace with a single

5 die, in one trial, is

6
; the probability of not throwing an ace is

6 the probability of throwing either an ace or a deuce, is &c.

that is,

(5.) Er. 3. If n balls a, b, c, d, &c. be thrown promiscuously into a bag, and a person draw out one of them, the probability that it will be a is-; the probability that it will either be a orb is 2.

n n.

(6.) Er. 3. The same supposition being made, if two balls be

2 drawn out, the probability that these will be a and 6 is

n.(n-1) n-1 For there are n. combinations of n things taken two and two together; and eacn of these is equally likely to be taken ; therefore

2 the probability that a and I will be taken is

n.(n-1)

2 (7.) Er. 3. If 6 white and 5 black balls be thrown promiscuously into a bag, and a person draw out one of them, the probability that

6 this will be a white ball is and the probability that it will be a

11 3 black ball is

11° (8.) From the bills of Mortality in different places, tables have been constructed which shew how many persons, upon an average, out of a certain number born, are left at the end of each year, to the extremity of life. From such tables, the probability of the continuance of a life, of any proposed age, is known.

(9.) Er. 1. To find the probability that an individual of a given age will live one year. Let A be the number, in the tables of the given age, B the number

B left at the end of the year ; then is the probability that the indivi.

A-B dual will live one year ; and the probability that he will die

А in that time. In Dr. Halley's Tables, out of 586 of the age of 22, 579 arrive at the age of 23 ? hence, the probability that an indi

579 83

7 1 vidual aged 22 will live one year is

nearly; and or 586 84

586'

84 nearly, is the probability that he will die in that time.

(10.) Er. 2. To find the probability that an individual of a given age will live any number of years. Let A be the number, in the tables, of the given age, B, C, D,..X,

B the number left at the end of 1, 2, 3,....t, years; then

the probability that the individual will live one year ;

the probabi

X lity that he will live two years; and the probability that he will

А A-B A-C A-X live & years. Also,

are the probabilities that he will die in 1, 2, 4, years.

These conclusions follow immediately from Art. 154.

(11.) If two events be independent of each other, and the probability that one will happen be and the probability that the other will happen, the probability that they will both happen is

For each of the m ways in which the first can happen or fail, may be combined with each of the n ways in which the other can happen or fail, and thus form mn combinations, and there is only one in which both can happen ; therefore the probability that this will be the case is I)

(12). Cor. 1. The probability that both do not happen is 1

For, the probability that they both liappen, together with the probability that they do not both happen, is certainty ; therefore, if from unity, the probability that they both happen be subtracted, the remainder is the probability that they do not both happen.

(13.) Cor. 2. The probability that they will both fail is (m-1)(n-1) For, the probability that the first will fail is

n-1 and the probability that the second will fail is

therefore the

(m-1) (.n-1)

probability that they will both fail is

ml n-1
X

. m

m n

(14.) Cor. 3. The probability that one will happen and the other fail is m+n2

For the probability that the first will happen and

n-1 the second tail is

Х and the probability that the first will