- (19).
- According to Theorem2.6.1 (Larsen's book pg59), we have
Let B be the event ``the program compiles on it's first
run'', and
be the event ``the program is written in COBOL'', and
be the event ``the program is written in FORTRAN,'' and we get
- (20).
- After we recall example 2.2.4, we let (x,y)
denote the
allocation scheme whereby x white chips and
y black chips are
put into urn I. And we get the probability of survive as following,
Notice here the probability of urn I or urn II is chosen is
same, 1/2. When
, we get the optimized
result, which is
14/19.
- (21).
- According to Theorem 2.6.1 (Larsen's book pg59), we have
Let R be the event ``R will win'',
be the event ``
will be nominated'', we get
- (23).
- Let
denote the actual
weather, sunny,
cloudy or rain, and let
denote the weather
forecasted to be sunny,cloudy or rain.
- (a).
-
- (b).
-
- (c).
-
- (24).
- Let B be the event ``the chip ultimately drawn from
urn I is red'', and RR be the event ``we draw a red chip
from urn
I and draw a red chip from urn II'', and RW be the event
``we
draw a red chip from urn I and draw a white chip from urn II'', and
WR be the event ``we draw a white chip from urn I and draw
a red
chip from urn II'', and WW be the event ``we draw a white
chip
>from urn I and draw a white chip from urn II'', and we get,
- (28).
- This can be solved by using a continuous analog of
Theorem2.6.1. If B is a event dependent on the value of
x, then
, where
f(x) is the
probability function for x. Let B be the event
that the second
point is at a distance < q from
x. f(x)=1, since point x
occurs everywhere in the interval (0,1) equally. Thus the probability
that the second point lies in an interval (a,b) is b-a. When point
x is in the interval (0,q), the points in the
interval (0,x+q)
are less than a distance of q from the point
x. When point x
is in the interval (q,1-q), the points in the interval
(x-q,x+q),
are less than a distance of q from the point
x. And when point
x is in the interval (1-q,1), the points in
the interval (x-q,1),
are less than a distance of q from the point
x. Hence we get
- (29).
-
,
. Since
the numerator of
is non-negative and
the denominator is
positive,
is non-negative, it
satisfies Axiom1.
, it satisfies
Axiom2. And if
and
are two mutually exclusive events,
Since
and
also are mutually
exclusive and P satisfies axiom 3,
Hence we get
Thus it also satisfies Axiom3.
- (30).
- Let B be the event ``the third chip drawn is
white'', and let RR be the event ``we first draw a red chip
then a
red chip'', and let RW be the event ``we first draw a red
chip then a
white chip'', and WR be the event ``we first draw a white chip
then a
red chip'', and let WW be the event ``we first draw a white
chip then a
white chip'', and we get,
- (32).
- From Theorem 2.3.6, we have
And from Theorem 2.3.1, we have
Thus we get following
- (33).
-
From theorem 2.3.6, we have
Hence we get
- (35).
- Consider the possible ways sets A and B could be related.
Case 1: A and B are mutually exclusive.
. Thus, the condition reduces to
.
This is equivalent to the condition,
Obviously, this is true if and only if
Case 2: A
B.
. Thus, the condition reduces to
.
This is equivalent to the condition,
Obviously, this is true if and only if
Case 3: A and B are not mutually exclusive and A is not a subset of B.
Again rewriting the condition in terms of the sizes of the sets involved we
get,
Note that
(I)
Since
and
(since A is not a
subset of B),
(II)
Since
and
(since A and B are
not mutually exclusive),
(III)
Now (I), (II), and (III) imply that
Which means that the condition is not true in this case.
In summation, the condition is true if and only if (B=S or B=
).
- (36).
- Base case:
.
Inductive step: Assume P(n) holds. show that
P(n+1) holds.
We have following assumption
Thus we get
- (37).
- Let W(n) and R(n) be the events that we pick repectively,
a white and red chip from the nth urn and let P[n] be the
probability
of picking white. Now,
Solving the recurrence we get :