Unformatted text preview: Problem 1.4.1 Solution
From the table we look to add all the disjoint events that contain H0 to express the
probability that a caller makes no handoffs as
In a similar fashion we can express the probability that a call is brief by
The probability that a call is long or makes at least two handoffs is Problem 1.4.5 Solution
Specifically, we will use Theorem 1.7(c) which states that for any events A and B,
.
To prove the union bound by induction, we first prove the theorem for the case of n = 2
events. In this case, by Theorem 1.7(c),
.
By the first axiom of probability, P[A1 ∩ A2] ≥ 0. Thus,
which proves the union bound for the case n = 2. Now we make our induction hypothesis
that the unionbound holds for any collection of n − 1 subsets. In this case, given subsets
A1, . . . , An, we define
・・・
By our induction hypothesis,
・・・ ・・・ This permits us to write
・・・
(by the union bound for n = 2)
・・・
・・・
which completes the inductive proof. Problem 1.4.6 Solution
(a) For convenience, let
unknowns and
fill the table as However, we are given a number of facts: . Using this shorthand, the six Other facts, such as
, can be derived from these facts. Thus, we have
four equations and six unknowns, choosing
and
will specify the other unknowns.
Unfortunately, arbitrary choices for either
or
will lead to negative values for the
other probabilities. In terms of
and p1, the other unknowns are Because the probabilities must be nonnegative, we see that Although there are an infinite number of solutions, three possible solutions are: and and (b) In terms of the , notation, the new facts are
extra facts uniquely specify the probabilities. In this case, Problem 4 and . These Problem 5 Problem 1.5.6 Solution
The problem statement yields the obvious facts that
and
. The
words“10% of the ticks that had either Lyme disease or HGE carried both diseases” can be
written as
(a) Since Thus,
P [LH] = 0.10P [L ∪ H] = 0.10 (P [L] + P [H] − P [LH])
Since P[L] = 0.16
P[H] = 0.10 (b) The conditional probability that a tick has HGE given that it has Lyme disease is Problem 7 Problem 1.6.1 Solution
This problems asks whether
and
can be independent events yet satisfy
? By
definition, events
and
are independent if and only if
. We can see
that if
, that is they are the same set, then
Thus, for A and B to be the same set and also independent,
There are two ways that this requirement can be satisfied:
•
implying • implying φ. Problem 1.6.3 Solution
(a) Since and are disjoint, A Venn diagram should convince you that . Since
c so that ,
. This implies It also follows that
(b) Events
and
are dependent since
(c) Since
and
are independent, .
. The next few items are a little trickier. From Venn diagrams, we see
It follows that
Using DeMorgan’s law, we have
(d) Since , and are independent. Problem 1.6.4 Solution
∅ (a) Since . To find , we can write Thus,
. Since
is a subset of
since
is a subset of
(b) The events A and B are dependent because
(c) Since and are independent . Furthermore, . So In addition,
we first observe that . To find , By De Morgan’s Law, . This implies Note that a second way to find
then
and
are independent. Thus is to use the fact that if C and D are independent, Finally, since and
are independent events,
.
(d) Note that we found
. We can also use the earlier results to show
(e) By Definition 1.7, events and are independent because Problem 1.6.5 Solution
For a sample space with equiprobable outcomes, consider the events Each event Ai has probability 1/2. Moreover, each pair of events is independent since
However, the three events A1,A2,A3 are not independent since Problem 1.7.4 Solution
The tree for this experiment is The probability that you guess correctly is Problem 1.7.5 Solution
The
is the probability that a person who has HIV tests negative for the disease. This
is referred to as a falsenegative result. The case where a person who does not have HIV but
tests positive for the disease, is called a falsepositive result and has probability P[+Hc].
Since the test is correct 99% of the time,
Now the probability that a person who has tested positive for HIV actually has the disease is We can use Bayes’ formula to evaluate these joint probabilities. Thus, even though the test is correct 99% of the time, the probability that a random person
who tests positive actually has HIV is less than 0.02. The reason this probability is so low is
that the a priori probability that a person has HIV is very small. Problem 1.7.10 Solution
The experiment ends as soon as a fish is caught. The tree resembles From the tree,
and
if no fish were caught on the previous . Finally, a fish is caught on the nth cast
casts. Thus, Problem 15 ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.
 Spring '08
 Ferry

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