HW2_Sol

HW2_Sol - Problem 1.4.1 Solution From the table we look to...

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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 hand-offs 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 hand-offs 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 union-bound 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 false-negative result. The case where a person who does not have HIV but tests positive for the disease, is called a false-positive 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.

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