RESULTS, HINTS, ETC., FOR THE EXERCISES ON

PROBABILITIES.

I.

2. The probabilities of a person alive at the beginning of each of the following six decades, not surviving 10 years:

Dr. Carpenter, the eminent physiologist, has remarked:—“It has been ascertained that from about the age of 18 to 28 the mortality is much greater in males than in females, being at its maximum at 25, when the viability (or probability of life) is only half of what it is at puberty. This fact is a very striking one, and shows most forcibly that the indulgence of the passions not only weakens the health, but in a great number of instances is the cause of a very premature death.”

II.

3. Suppose the 3 the number of points,

then the chance of throwing 3 in the first throw is f.
And in the second throw the chance of failure is .
Therefore the required chance is

5$ 125
6216'

and of success 1

125 91

=

216 216

5. In each of his three throws A may throw 1, 2, 3, 4, 5, or 6 points, and the number 8 can be made up by the following throws:

1, 1, 6; 2, 2, 4; 3, 3, 2; 1, 2, 5; 1, 3, 4. Next B in three throws can make up 7 points by the throws

1, 1, 5; 2, 2, 3; 3, 3, 1; 1, 2, 4.

6. One throw gives 4 chances, 3 unfavourable, and 1 favourable.
Two throws give 16
Three

9

64

27

Therefore two throws give the odds against, 9 to 7;

three throws give the odds for, 37 to 27, greater than an even chance by 12

7. With one throw there is 1 chance favourable and With two throws there are 11 chances

5 unfavourable.

25

With three throws the odds against are 125 to 91;

but with four throws the odds for are 671 to 625.

There is less than an even chance in three throws, but greater than an even chance in four throws, and still greater than an even chance in five throws, &c.

8. Since in one throw there is one chance of succeeding and five chances of failing,

I

Suppose x the number of throws, then probability of failing in a throws. But as the wager is the same, failing and succeeding, this probability of success and failure will be represented by 1,.., taking the logarithms of this equa

tion x{log 5-log 6} = -log 2, .. x=;

=

log 2 •20103 log 6-log 507918

=

=382 nearly; that is,

the chance is greater than 3, but less than 4, and nearer 4 than 3. In 4 throws there is also an equal chance of throwing any one of the other five faces of the die.

There is an even chance of throwing two aces with two dice when

which x 24.605 nearly.

9. First suppose one ace to be cast, then setting aside one die, there remain 9 dice, of each of which only 5 points must be taken.

The number of combinations of these will be 5o, which when combined with all the aces in the 10 dice, will give 10.5°; and the number of the combinations is 610. 10.5° 19531250 1 .. The probability of casting one ace is 610 60466176 3 Next if two aces be cast, then setting aside two dice there remain 8 dice, of each

of which as before 5 points are to be taken.

=

nearly.

The number of combinations of these will be 58, which when combined with all the aces of the 10 dice taken 2 at a time will give .5, and the whole number of

combinations is 610.

10.9
1.2

the number of favourable chances of 3 aces with 10 dice, and

57 is the number of combinations of the other 5 points.

10.9.8 57 9375000 2 nearly. 1.2.3 10 60466176 13

=

Therefore the probability of casting three aces is 10. First, the chance of the tetrahedron falling on some one of four faces marked 1, 2, 3, 4, is -1.

The chance of the cube falling on some one of four faces marked 1, 2, 3, 4, is =3.

The chance of the octahedron falling on some one of four faces marked 1, 2, 3, 4, is =.

The chance of the dodecahedron falling on some one of four faces marked 1, 2, 3, 4, is 2}.

.. The chance of all four falling on some one of four faces is Ixxx!=& The same result will appear from the consideration.

All the possible cases that can occur with the four polyhedrons is

4X6X8X12=2304.

And the cases of the four specified numbers is 4 × 4×4×4=256. .. The probability required is

11.10.9.8
1.2.3.4

5.4.3
1.2.3

.

III.

=all the combinations of the black and white balls,

-all the combinations of the 5 white balls, 3 at a time.

5.4.3 11.10.9.8 10 1
1.2.3 1.2.3.4 330 33

= 630 the number of pairs which can be drawn.

10 x 12 120 pairs, one white and one black.
10×14=140 pairs, one white and one red.
12×14=168 pairs, one black and one red.

And since there are 120 favourable chances out of 630 chances, the probability of

120 4 success is and the probability of failure is 1– 630 21 against are as 17 to 4.

Next C1. C1. C1. 3C1=3.3.3.3=81, the number of favourable combinations of the 12 things 4 at a time.

9. Here is a case of an event being dependent on another event. exchange the urn A contains 1 black and 2 white balls; and B, 1 white and 2 black balls.

In the second exchange the probability of a white ball taken from A is

2

also the

After the first

probability of a black ball taken from B is also 2:

2 2 4

these events will happen is X
3 3 9

: and the probability that both

On this hypothesis, that A now contains 1 white and 2 black balls, and B contains 1 black and 2 white balls,

1

3'

In the third exchange, the probability of a white ball being taken from A is

and a black ball from B is also : and the probability that both these events will

Now since the probability in the second exchange is that A contains 2 black and B 2 white balls. Also since the probability in the third exchange is that A contains 1 black and B 1 white ball,

Therefore the probability of 3 black balls in A and 3 white in B is

4.1

9 81

Hence

and 11C4:

=

is the probability that the 4 balls drawn

11.16 and C2

• 1.2

1.2

is the number of ways in which 4 balls can be drawn,

11C.C2 11.10.2.3.x. (x-1)

=

(x+11)(x+10)(x+9)(x+8)'

+C
+1 C1
11.10.9.8

=

11.10.2.3.x. (x-1)

(x+11)(x+10)(x+9)(x+8) ̄(x+11)(x+10)(x+9)(x+8)

or 12=x(x-1), x= = 4 and

12. The number 10 can be made up in four ways by taking three numbers1, 4, 5; 2, 3, 5; 3, 3, 4; 4, 4, 2.

13. First find the number of combinations of all the 8 numbers taken 4 at a time. Next all the number of combinations of four of the numbers whose sum is 17, bearing in mind that every one of the 8 numbers is likely to be drawn each time.

IV.

1. Since A's skill is to E's as 3 to 1. This implies that A wins 3 games while B wins 1... A's chance of winning one game is, and B's 1.

6. The chance of the correctness of A's affirmation is, and of his negation . And the chance of B's affirmation, of his negation is 4.

... The chance of A and B's affirmation is ×=}, and
of A's and B's negation is 4×7=24

And these two agreements are exclusive.

.. The chance of one or the other is +14.

That is they agree in 16 out of 24 of their statements.

8. The chance of a white ball being drawn is; of its not being drawn f. The chance of A speaking truth is, and of his failure 1.

9. Here the chance of winning is equal to the chance of losing a game, and therefore is equal to . And each of the three simple chances is . Therefore in each of the three cases the chance required is ×× }=}. 10. By reference to the general formula, Art. 5. n=6, r=2, and in this case the a®+6a5b+15a+b2 chance is where a 1 and b=20, its numerical value is

(a+b)

6121

216

1. There are 10+5+3+2+1=21 coins, and two of them are to be drawn. The chance of any two being drawn is, and the expectation is

2. The chance for 4s. is; for 8s. 4, and for 5s. .

(£10 14s. 1d.).

.. The value of the expectation is

12+48 +25 85
14
14

shillings.

4. Here 4+6=10, and is the equal chance for drawing each prize.

The value of the expectation is

=

10

3+5+8+10 26
2 shillings.
10

5. The bag contains 9 coins; if 3 be drawn, the chance is, or of the value of

the coins; that is of 107 shillings, or £1 15s. 8d.

6. The price of a ticket will be the sum of the expectations for each prize.

There are 4 tickets out of 100 for a prize of £100;

Similarly, the value of the expectation of a prize of £50 is

And the value of the expectation of a prize of £5 is

5

The sum of the expectations is £10, the value of a ticket.