Find John A Gubner solutions at now. Below are Chegg supported textbooks by John A Gubner. Select a textbook to see worked-out Solutions. Solutions Manual forProbability and Random Processes for Electrical and Computer Engineers John A. Gubner Univer. Solutions Manual for Probability and Random Processes for Electrical and Computer Engineers John A. Gubner University of Wisconsin–Madison File.
|Published (Last):||3 April 2005|
|PDF File Size:||2.79 Mb|
|ePub File Size:||5.25 Mb|
|Price:||Free* [*Free Regsitration Required]|
Skip to main content. Log In Sign Up. But f x must be one of the values y1. We must show that B is countable.
Otherwise, there is at least one element S of B in A, say ak. We must show that B is uncountable. We prove this by solutionw. Suppose that B is countable. Then by the previous problem, A is countable, contradicting the assumption that A is uncountable. Suppose A is countable and B is uncountable.
Using only the numbers 1, 2, 3, 4, 5, 6, consider how many ways there are to write the following numbers: We must check the four axioms of a probability measure. Third, for disjoint Chapter 1 Problem Solutions 7 For arbitrary solutiosn Fnlet Aolutions be as in the preceding problem. For arbitrary events Gnput Fn: Following the hint, we put Gn: In this problem, the probability solutionw an interval is its length.
Second, since A1. Hence, the collection is closed under complementation. Our proof is by contradiction: Now, any set in A must belong to some An. Then A contains each union of the form [ Ai.
T c First note that by part aA: Chapter 1 Problem Solutions 11 Here is an example. Then E does not belong to A since neither E nor E c the gunber integers is a finite set. First suppose that A1. In the second case, suppose that some A cj is finite.
Errata for Probability and Random Processes for Electrical and Computer Engineers
To see this, put Ai: Let A denote the collection of all subsets A such that either A is countable or A c is countable.
First, the empty set is countable. There are two cases. Third, let A1S, A2.
There are two cases to consider. If all An are countable, then n An is also countable by an earlier problem. Let I denote the collection of open intervals, and let O denote the collection of open sets.
Recall that in the problem statement, it was shown that every open set U can be written as a countable union of open intervals. Let O denote the event that a cell is overloaded, and let B denote the event that a call is blocked. Let H denote the event that a student does the homework, and let E denote the event that a student passes the exam.
Let F denote the event that a patient receives a flu shot. Without loss of generality, let 1 and 2 correspond to the two defective chips. To prove this, we construct a sample space and probability measure and compute the desired conditional probability. Here i and j are the chips taken by the friend, and k is the chip that you test.
We again take 1 and 2 to be the defective chips. The 10 possibilities for i and j are 12 13 14 15 23 24 25 34 35 45 For each pair in the above table, there are three possible values of k: For the probability measure we take P A: For example, if the friend takes chips 1 and 2, then from the second table, k has to be 3 or 4 or 5; i.
The event that the friend takes two chips is then T: Now the event that you test a defective chip is D: Of the above intersections, the first six intersections are singleton sets, and the last three are pairs.
If two disjoint events both have positive probability, then they cannot be independent. Let W denote the event that the decoder outputs the wrong message. Of course, W c is the event that the decoder outputs the correct message. Now, W c occurs if only the first bit is flipped, or only the second bit is flipped, or only the third bit is flipped, or if no bits are flipped.
Denote these disjoint events by FFFand Frespectively. Let L denote the event that the left airbag works properly, and let R denote the event that the right airbag works properly.
Let Wi denote the event that you win on your ith play of the lottery. Let A denote the event that Anne catches no fish, and let B denote the event that Betty catches no fish. We show that A, B, and C are mutually independent. We show that the probability of the complementary event is zero. There are 2n n-bit numbers. In order that the first header packet to be received is the 10th packet to arrive, the first 9 packets to be received must come from the 96 data packets, the 10th packet must come from the 4 header packets, and the remaining 90 packets can be in any order.
More specifically, there are 96 possibilities for the first packet, 95 for the second. Suppose the player chooses distinct digits wxyz. The player wins if any of the 4!
In the first case, since the prizes are different, order is important.
There are 52 14 possible hands. All five cards are of the same suit if and only if they are all spades or all hearts or all diamonds or all clubs. These are four disjoint events. Hence, solitions answer is four times the probability of getting all spades: There are k1Two apples and three carrots corresponds to 0, 0, 1, 1, 0, 0, 0. Five apples corre- sponds to 0, 0, 0, 0, 0, 1, 1. The possible values of X are 0, 1, 4, 9, Gjbner X1X2X3 be the random digits of the drawing.
Chapter 2 Problem Solutions 23 Let Xi be the price of stock i, which is a geometric0 p random variable. Let Xk denote the number of coins in the pocket of the kth student.
Then the Xk are independent and is uniformly distributed from 0 to 20; i. Hence, X and Y are independent. Here is a script: The straightforward approach is to put f c: The two sketches are: The right-hand side is easy: The left-hand side is more work: So the bound is a little more than twice the value of the probability.
The true probability is 0. As discussed in the text, for uncorrelated random variables, the variance of the sum is the sum of the variances.
Since independent random variables are uncorrelated, the same results holds for them too. Next, since the Xk are i. The remaining sum is obviously nonnegative. Assume the Xi gybner independent.
Z Since Z is the sum of i. Let Xi be i. Chapter 3 Problem Solutions 35 If the Xi are i. Hence, they are uncorrelated. We have from the example that with p: