13. The sum of the favorable and unfavorable events equal the No. of combinations of 25 things, 10 at a time

The combinations of 5 black balls, 3 at a time

20 white balls, 7

= .........

.. the sum of the favorable events =

.. the probability required

14. The No. of combinations of 52 things, 13 at a time

The No. of combinations of 48 things, 12 at a time

2.3 ... 12

With each of the latter set, only one of the 4 aces may be combined; .. chance first required

=

39.38.37. (13 × 4)_ 9139 703-7 nearly.

=

=

52.51.50.49 20825 1602

Also, chance of dealing one ace to each person

16

16.

Chance of dipping into the 1st or 2nd urn=

and the chance of drawing white from 2nd urn ==

18. The even chance of throwing an ace in x times

19. Let a, b be the numbers of white and black balls respectively; n the number of bags.

The probability of drawing a white ball from each bag

black

a

=

a+b'

The proby. that the n balls drawn will all be white

The most probable number corresponds with that expression in the above series which is greatest;

Sincer is an integer, it must = 2;

.. the most probable number of white balls drawn = n − r = 10.

20. The number of ways in which this can be done is expressed by the coefficient of 16 in the expansion of (x + x2 + x3 + x2 + x5 +x6)+,

which coefficient, found by the Multinomial Theorem, is

6 +24+ 12 + 12 + 12 + 1 + 12 + 24 + 4 + 6 +12= 125; but there are 64 1296 possible throws, hence the required

=

chance =

125 1296'

16.

Here we have to express 1719 by means of the terms of the series 1, 2, 22, 23, &c.; and to transform 1719 into the binary scale, we have

2) 1719
2859......I
2)429......I
2)214......I
2) 107......0
2)53......I
2) 26......I
2) 13......0
2)6......I

2)3......0

I......I

whence the number 1719 is equivalent
to IIOIOIIOIII in the binary scale, -
which is expressed by 1+2+22 + 24
+25 + 27+ 29 +2°; and therefore it
will be necessary to select the weights.
I lb., 2lbs., 22lbs., 24lbs., 25 lbs., 27lbs.,
29 lbs., and 21o lbs.

18. Let αo, a, a, a,, &c. be the digits, so that

N=α。 + 10α, + 102α2 + 103α, + &c.

of which the general term is 10"am; then with the same digits differently arranged to make N', the general term containing am will be 10"am; and therefore the general term of N~ N' will be

(10” — 10") am = 10" (10m—” — 1) am if m>n,

or = 10m (10”—m — 1) am if n > m;

since this term is divisible by io - I;

.. every term in N~ N' is divisible by 9.

LOGARITHMS.

Ex. 70.

1. log 5'4=3 log 3 + log 2 - log 10 = 1'431363

+301030-1='732939.

log 17.5= log 7+ 2 log 15-2 log 3 - log 10

= ·845098+2*352182-954242 – I

=

= 1*243038.