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| − | Solutions | + | =Solution= |
| + | 2.62 <br> | ||
| + | <math> P(A|B)= \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} =1 \, (since\ B \subset A).</math> <br> | ||
| + | <math> P(B|A)= \frac{P(A \cap B)}{P(A)} =\frac{P(B)}{P(A)} = \frac{2}{7}.</math> | ||
| + | |||
| + | Solutions for other problems can be viewed at the open online source: <br> | ||
| + | http://www.athenasc.com/probbook.html | ||
| + | |||
| + | Feel free to ask questions if there's any. -TA | ||
=Questions/comments/Discussion== | =Questions/comments/Discussion== | ||
Latest revision as of 05:23, 31 January 2013
Homework 2 Discussion and Solutions, ECE302, Spring 2013, Prof. Boutin
- Problem 2.62 from the textbook: Probability, Statistics, and Random Processes for Electrical Engineering, 3rd Edition, by Alberto Leon-Garcia, Pearson Education, Inc., 2008.
- Problems 14, 15, 16, 17, 18, 24, 25 30, 31, 32, 33 34 from Chapter 1 of "Introduction to Probability," by Dimitri P. Bertsekas and John N. Tsitsiklis. Athena Scientific, Belmont, Massachusetts, 2008.
Solution
2.62
$ P(A|B)= \frac{P(A \cap B)}{P(B)} = \frac{P(B)}{P(B)} =1 \, (since\ B \subset A). $
$ P(B|A)= \frac{P(A \cap B)}{P(A)} =\frac{P(B)}{P(A)} = \frac{2}{7}. $
Solutions for other problems can be viewed at the open online source:
http://www.athenasc.com/probbook.html
Feel free to ask questions if there's any. -TA
Questions/comments/Discussion=
- Write a question here.
- answer here.
