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Stat8001 hw1, due Sept 24
th
: Solve exercise problems: 1.5, 1.7, 1.13,
and 1.
1.5, p.
a) The event A B C is such that {a U.S. birth results in identical twin females}
b)
P A B C
1.7, p.
a)
Example 1.2.7 (on p. 8) had derived the formula: P (scoring i points) =
2 2
2
(6-i) - (5-i)
However, in Example 1.2.7, the probability was set = 1 for the dartboard itself. In this question
however, we are to assume the probability is 1 for the entire wall. Thus we must adjust the
above formula by multiplying by the ratio of the dartboard’s area / the entire wall’s area:
P (scoring i points) =
2 2 2
2
(6-i) - (5-i)
r
A
where r is the (unknown) radius of the dartboard.
Note however, that this formula only works for cases in which i = 1, 2, 3, 4 or 5. It will not work
for i=0 because that is outside the circle shown in Figure 1.2.1 (see p. 9). In the case that i=0,
the probability of scoring i points (i.e. 0 points) is simply related to the ratio of A to
2
r. As
2
r approaches 0, the probability of scoring i points (i.e. 0 points) approaches 1. The
probability of scoring i points (i.e. 0 points) is simply the difference between the probability of
hitting the wall (1) – (the size of the board relative to the size of the wall):
P (scoring i points) = 1 -
2
r
A
where i = 0.
b)
By definition (Definition 1.3.2, p.20) of Conditional Probability:
P (scoring i points | board is hit) =
the probability of scoring i points the probability the board is hit
the probability the board is hit
We know that
the probability of scoring i points is
2 2
2
(6-i) - (5-i)
and
the probability the board is hit is:
2
r
A
P (scoring i points | board is hit) =
2 2 2
2
2
(6-i) - (5-i)
r
A
r
A
2 2
2
(6-i) - (5-i)
Of course, if board is hit, then i=1, 2, 3, 4, or 5.
1.13, p.
By definition (Definition 1.1.5 on p.5), two events A and B are “disjoint,” or
“mutually exclusive,” if there is no intersection between the events i.e. if
A B
. For this to occur, when finding
P A ( B )
, it would be =
P A ( ) P B ( ) P A ( B )
which simplifies to =
P A ( ) P B ( )
in the case of disjoint events because
P A ( B ) 0
. (See the
Axiom of Finite Additivity, p.9.) And of course the sum of
P A ( ) P B ( )
must be 1 because 1 is
the maximum, or “ceiling,” probability of the sample space. However, when we add
P A ( ) P B ( )
( ) [1 ( )] [1 ]
C
P A P B
, we get a result that is
greater than 1.
A and B cannot be disjoint.
Another way to approach this problem is by noticing that if A and B were to
be disjoint, A would need to be part of the sample space outside of B i.e.
C
A B
. How would we test this? Well, if
C
A B
then the maximum probability A could have
is the probability that
C
B
has. I.e.,
C
P A P B
. (This is a conclusion I made based on
Theorem 1.2.9c on p.10.) In this case,
P A
and
C
P B
so obviously our
P A ( )
is not
our ( )
C
P B , which again proves that A and B cannot be disjoint.
1.14, p.
If the sample space S contains only n=1 element—e.g. S = {a}, the total
number of subsets is simply the null set, and the set containing that 1
element i.e. {
, a }.
There are a total of 2 subsets, or 2
1
subsets. If we add
P A B
P A B
P B
What happens here is that B becomes the entire sample space. I.e. our original sample space has
been updated to B. We don’t care about situations in which B is not occurring. So P(B) is the be
all and end all, you might say. The
P B ( | B ) 1
because the probability of B occurring when B is
occurring is 100%. If A contains all of the different data points of B, then you would conclude that the
P A ( B )
is nothing but
P B ( )
so
P A B P B
P A B
P B P B
. In other words, in such a case,
we are calibrating all occurrences with respect to B. If A and B intersect at all points of B, then
P A B ( | )
will include (the probability all those point of B, that are also in A) / (the probability of B), which equals
(the probability of B) / (the probability of B), which equals 1.
Now, in another extreme case, imagine that A and B are totally disjoint. In such a case, there is
no intersection between A and B, so
P A ( B ) 0
, so
P A B
P A B
P B P B
The same could be said for
P B ( | A )
in this case:
P B A
P B A
P A P A
Thus, in this case,
P A B ( | ) P B ( | A )
Defn 1.41, p.
A Random Variable, RV, is a function from a sample space S into the Real numbers. For
example, if you take a poll of 50 Congressmen who are going to vote either for cutting children’s
health care spending vs. leaving it at current levels, you could summarize the resultant data by
recording a “1” for “leaving” and “0” for cutting. This would give you a sample space that has
50
= 1125899906842620 elements, each of which is an ordered string of 1s and 0s of length 50.
The order of the votes probably doesn’t matter though; we’d probably just want the total number
of “1s” in the sample. To get this, we’d simply define a random variable X, which is going to be
an integer between 0 and 50. Thus, the sample space of X is
Defn 1.5.1, p.
No matter what a Random Variable is, it is associated with a
The Cumulative Distribution Function, or CDF, of a random variable (RV) X, is denoted by
X
F x , and is defined as
X X
F x P X x , for all
x
.
Defn 1.5.7, p.
