Probability for Engineering Applications
Mid Term Exam #1, October 20th, 2003
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Do This First:
1. Make sure that you have all the pages of the exam book. 2. Fill in the abo

11/21/03 1:01 PM
HW 13page 1
Probability for Engineering Applications
SOLUTION: Assignment #13
Two random variables.
1. (2 pts) Let X be the input to a communications channel (cf 4.17). X takes on the values +1 and 1 with equal probability. The o

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HW12, page 2
Joint empirical CDF of X and Y
1 0.8 0.6 0.4 0.2 0
19 ACTUAL 20 21 0 22 0.1 23 0.2 24 0.3 25 0.4 26 0.5 27 0.6 28 0.7 29 0.8 30 0.9 31 1 32 1.1 3

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SOLUTION
Probability for Engineering Applications
Assignment #11
1. (3 pt) For the two-dimensional random variable Z= (X,Y) sketch the region of the plane corresponding to the following events and state whether the events are in product form. (cf

10/31/03 9:01 AM
HW10 solution page 1
Probability for Engineering Applications
Solution to Assignment #10
1. (3 pts) The lifetime T (years) of a light bulb has pdf a. Find the reliability function and the MTTF of the bulb. R(t) = P(T>t) = 1 =1 fT(

Solutions to Homework 08
1. Y = eX . We denote the pdf of X by fX (x) and its cdf by FX (x). Also, Y = eX X = ln Y . a. Note rst that for all values of X, i.e. < x < , Y > 0. Thus, P [Y 0] = 0. For Y > 0, FY (y) = = = = Therefore FY (y) = Now, fY

Solutions to Homework 07
1. a. In this case we have transmitted a 0 (i.e., v = 1). The receiver makes an error if the received signal is greater than 0 (i.e., if v + N 0). Then P [error|v = 1] = = = = = b. Similarly, if a 1 is sent P [error|v = 1] =

Solutions to Homework 06
1. a. From the gure, fX (x) = Now, 1= =
a
0 c 1
|x| a
|x| > a |x| a
fX (x)dx
|x| dx a a a x c 1 =2 dx a 0 a x2 = 2c x 2a 0 = ac c 1 1 a b. FX (x) = 0 for x < a and FX (x) = 1 for x > a. For a x 0, |x| = x. Then c=

Solutions to Homework 05
1. 4 a. P [ace in rst draw] = 52 . b. Suppose the rst draw is seen to be an ace. Then, we now have 3 aces remaining in the 51 cards. Thus, 3 P [ace in second draw | ace in rst draw] = 51 If the the rst draw is not an ace, we

Solutions to Homework 04
1. a. P [A B] = P [A] + P [B] P [A B] = P [A] + P [B] P [A]P [B]. b. P [A B] = P [A] + P [B] P [A B] = P [A] + P [B]. (since A B = P [A B] = 0) 2. a. This is a case of the Binomial probability law. P [k errors] = n

Solutions to Homework 03
[1] The number of ways of picking 20 raccoons out of N is N 20 The number of ways of picking 5 tagged raccoons out of 10 and 15 untagged raccoons out of N-10 is 10 N 10 5 15 The probability of picking 5 tagged out of 20 is 1

Solutions to Homework 02
1. We are interested in the shaded region shown in Figure 1. The shaded are on the left corresponds to A B C while that on the right is AC B. Also, the unshaded area common to both A and B is A B. Now, note that A = (A B

SOLUTION PEA Final Examination Thursday December 16, 2002
All six questions required for full grade. Please give explicit numerical answers wherever possible, on the same page as the question.
Q1 Q2 Q3 Q4 Q5 Q6 TOTAL
15 15 20 15 15 20 100
12/10/03

11/24/034:41 PM
Probability for Engineering Applications
Mid Term Exam #2, November 13th, 2003
Last name of student
Student ID number
First name of student
Email address
Do This First:
1. Make sure that you have all the pages of the exam book.

Probability for Engineering Applications
SOLUTION Assignment #15
The Central Limit Theorem has infinite applications. Here are four short examples. 1. (2 pts) A fair coin is tossed 1000 times. Estimate the probability that the number of heads is bet