# HW05 - University of Illinois Fall 2009 ECE 313 Problem Set...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 5: Solutions Conditional Probability, Law of Total Probability, Bayes’ Formula 1. [Picking a random subset] (a) There are n k different subsets of size k , and since each is equally likely to be drawn, the probability of drawing a specific subset is just n k- 1 = k × ( k- 1) ×···× 2 × 1 n × ( n- 1) · · · × ( n- k + 1) . (b) There are k “desirable” objects among the set of n objects and if we draw one at random, the probability that it is a “desirable” object is k/n , that is, P ( A 1 ) = k/n . If the event A 1 has occurred, then there are only k- 1 “desirable” objects left among the n- 1. Thus, P ( A 2 | A 1 ) = ( k- 1) / ( n- 1). If both A 1 and A 2 have occurred, there are k- 2 “desirable” objects left among the n- 2 and so P ( A 3 | A 1 A 2 ) = ( k- 2) / ( n- 2) and so on. More generally, P ( A i | A 1 A 2 · · · A i- 1 ) = ( k- ( i- 1)) / ( n- ( i- 1)) since i- 1 “desirable objects” have been drawn already. Finally, P ( A 1 A 2 · · · A k ) = P ( A 1 ) P ( A 2 | A 1 ) · · · P ( A k | A 1 A 2 · · · A k- 1 ) = k × ( k- 1) ×···× 2 × 1 n × ( n- 1) · · · × ( n- k + 1) . 2. [The beginning of spring training] Since the pitcher can only pitch fast balls, curve balls, and sliders, we have that P ( F ) + P ( C ) + P ( S ) = 1, and since P ( C ) = 2 P ( F ), we conclude that P ( S ) = 1- 3 P ( F ). The law of total probability gives P ( H ) = 1 4 = P ( H | F ) P ( F ) + P ( H | C ) P ( C ) + P ( H | S ) P ( S ) = P ( H | F ) P ( F ) + P ( H | C ) · 2 P ( F ) + P ( H | S )(1- 3 P ( F )) = 2 5 P ( F ) + 1 4 · 2 P ( F ) + 1 6 (1- 3 P ( F )) from which we get that P ( F ) = 5 24 ,P ( C ) = 10 24 , and P ( S ) = 9 24 . 3. [Exercises in Conditional Probability] (a) P ( A | B c ) = 1- P ( A c | B c ) = 0 . 6. P ( A ) = P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) = 0 . 3 × . 7 + 0 . 6 × . 3 = 0 . 39. P ( B | A ) = P ( A | B ) P ( B ) /P ( A ) = 0 . 3 × . 7 / . 39 = 7 / 13....
View Full Document

## This note was uploaded on 10/25/2010 for the course ECE 313 taught by Professor Milenkovic,o during the Spring '08 term at University of Illinois, Urbana Champaign.

### Page1 / 3

HW05 - University of Illinois Fall 2009 ECE 313 Problem Set...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online