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EXAMPLES. XXXVII. b.

1. Find the number of permutations which can be made from all the letters of the words, (1) irresistible,

(2) phenomenon,

(3) tittle-tattle. 2. How many different numbers can be formed by using the seven digits 2, 3, 4, 3, 3, 1, 2? How many with the digits 2, 3, 4, 3, 3, 0, 2?

3. How many words can be formed from the letters of the word Simoom, so that vowels and consonants occur alternately in each word ?

4. A telegraph has 5 arms and each arm has 4 distinct positions, including the position of rest: find the total number of signals that can be made.

5. In how many ways can n things be given to m persons, when there is no restriction as to the number of things each may receive?

6. How many different arrangements can be made out of the letters of the expression a5b3cl when written at full length ?

7. There are four copies each of 3 different volumes; find the number of ways in which they can be arranged on one shelf.

8. In how many ways can 6 persons form a ring? Find the number of ways in which 4 gentlemen and 4 ladies can sit at a round table so that no two gentlemen sit together.

9. In how many ways can a word of 4 letters be made out of the letters a, b, e, c, d, o, when there is no restriction as to the number of times a letter is repeated in each word ?

10. How many arrangements can be made out of the letters of the word Toulouse, so that the consonants occupy the first, fourth, and seventh places?

11. A boat's crew consists of eight men of whom one can only row on bow side and one only on stroke side: in how many ways can the crew be arranged ? 12. Show that

n+1Cr=Cr+"Cr-1:

CHAPTER XXXVIII.

PROBABILITY (CHANCE).

a

a+b'

376. DEFINITION. If an event can happen in a ways and fail in b ways, and each of these ways is equally likely, the probability, or the chance, of its happening is and of its

6 failing is

a+b For instance, if in a lottery there are 7 prizes and 25 blanks, the chance that a person holding 1 ticket will win a prize is 7

25 and his chance of not.winning is

32 Instead of saying that the chance of the happening of an event is it is sometimes stated that the odds are a to b in favor of the event, or b to a against the event.

Thus in the above the odds are seven to twenty-five in favor of the drawing of a prize, and twenty-five to seven against

a

a+b'

success.

377. The reason for the mathematical definition of probability may be made clear by the following considerations :

If an event can happen in a ways and fail to happen in b ways, and all these ways are equally likely, we can assert that the chance of its happening is to the chance of its failing as a to b. Thus if the chance of its happening is represented by ka, where k is an undetermined constant, then the chance of its failing will be represented by kb.

chance of happening + chance of failing =k(a+b). Now the event is certain to happen or to fail; therefore the sum of the chances of happening and failing must represent certainty. If therefore we agree to take certainty as our unit, we have

1
1=k(a+b), or k=
.. the chance that the event will happen is

a+

6 and the chance that the event will not happen is

a+b

a+s

a

с

or

COR. If p is the probability of the happening of an event, the probability of its not happening is 1-p.

378. The definition of probability in Art. 376 may be given in a slightly different form which is sometimes useful. If c is the total number of cases, each being equally likely to occur, and of these a are favorable to the event, then the probability that the event will happen is, and the probability that it will not happen is 1–4.

Example 1. (a) From a bag containing 4 white and 5 black balls a man draws a single ball at random. What is the chance that it is black ?

A black ball can be drawn in 5 ways, since any one of the 5 black balls may be drawn. In the same way any one of the 4 white balls may be drawn.

5 5 Hence the chance of drawing a black ball is

4+5' 9 (6) Suppose the man draws 3 balls at random. What are the odds against these being all black ?

The total number of ways in which 3 balls can be drawn is 'C3, and the total number of ways of drawing 3 black balls is 5C:; therefore the chance of drawing 3 black balls

5 Cz

5.4.3 5

9.8.7 42 Thus the odds against the event are 37 to 5.

Example 2. A has 3 shares in a lottery in which there are 3 prizes and 6 blanks ; B has 1 share in a lottery in which there is 1 prize and 2 blanks. Show that A's chance of success is to B's as 16 to 7. A may draw 3 prizes in 1 way;

3.2 he may draw 2 prizes and 1 blank in

1.2

6.5
he may draw 1 prize and 2 blanks in 3 x ways;

1.2 the sum of these numbers is 64, which is the number of ways in

9.8.7 which A can win a prize. Also he can draw 3 tickets in

1.2.3

=

[ocr errors]

x 6 ways;

or

84 ways;

64 16 therefore A's chance of success =

84 21

B's chance of success is clearly 3;

16 1 therefore A's chance : B's chance

21 3

=16:7. Or we might have reasoned thus: A will get all blanks in 6.5.4

20 5 or 20 ways; the chance of which is or 1.2.3

84' 21

5 16 therefore A's chance of success =1

21

[ocr errors]

21°

379. From the examples given it will be seen that the solution of the easier kinds of questions in Probability requires nothing more than a knowledge of the definition of Probability, and the application of the laws of Permutations and Combinations.

EXAMPLES XXXVIII.

1. A bag contains 5 white, 7 black, and 4 red balls; find the chance of drawing: (a) One white ball; (6) Two white balls ; (c) Three white balls; (d) One ball of each color; (e) One white, two black, and three red balls.

2. If four coins are tossed, find the chance that there should be two heads and two tails.

3. One of two events must happen: given that the chance of the one is two-thirds that of the other, find the odds in favor of the other.

4. Thirteen persons take their places at a round table. Show that it is five to one against two particular persons sitting together.

5. There are three events A, B, C, one of which must, and only one can, happen; the odds are 8 to 3 against A, 5 to 2 against B. Find the odds against C.

6. A has 3 shares in a lottery containing 3 prizes and 9 blanks; B has 2 shares in a lottery containing 2 prizes and 6 blanks. Compare their chances of success.

7. There are three works, one consisting of 3 volumes, one of 4, and the other of 1 volume. They are placed on a shelf at random. Prove that the chance that volumes of the same works are all together is Tảo

8. The letters forming the word Clifton are placed at random in a row. What is the chance that the two vowels come together?

9. In a hand at whist what is the chance that the four kings are held by a specified player ?

10. There are 4 dollars and 3 half-dollars placed at random in a line. Show that the chance of the extreme coins being both halfdollars is .

CHAPTER XXXIX.

BINOMIAL THEOREM.

380. IT may be shewn by actual multiplication that
(x+a) (x+2)(x+c) (x+d)
== x++(a+b+c+d) 2-3 +(ab+ac+ad+bc+bd+cd) .2.2
+(abc+abd+acd+bcd) x+abcd .......

(1). We

e may, however, write down this result by inspection; for the complete product consists of the sum of a number of partial products each of which is formed by multiplying together four letters, one being taken from each of the four factors. If we examine the way in which the various partial products are formed, we see that

(1) the term 24 is formed by taking the letter x out of each of the factors.

(2) the terms involving 23 are formed by taking the letter x out of any three factors, in every way possible, and one of the letters a, b, c, d out of the remaining factor.

(3) the terms involving 22 are formed by taking the letter s out of any two factors, in every way possible, and two of the letters a, b, c, d out of the remaining factors.

(4) the terms involving x are formed by taking the letter a out of any one factor, and three of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters

a, b, c, d.

Example. Find the value of (c – 2)(x+3)(x – 5) (x + 9). The product =x4+ (-2+3 – 5+9) 2c +(-- 6+10 - 18 - 15+27 -- 45) ac3

+(30 – 54+90 – 135) x + 270 =x4+ 5x3 — 47x2 – 69.x+ 270.

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