Solution to HW 2 - Solution to EE351K HW assignment 2 1 ABC...

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

View Full Document Right Arrow Icon
Solution to EE351K HW assignment 2 1 ABC AB C C ABC C (0.04) AB C C C (0.16) A C BC (0.05) A C B C C (0.05) A C BC C (0) A C B C C C (0.2) Since ( ) 0.1 C P A C = , we know that ( ) 0.1 ( ) 0.05 C C C P A B C P A BC = - = . Since P(A)=0.7, we know ( ) 0.3 C P A = , hence ( ) 0.3 0.05 0.05 0.2 0 C C P A BC = - - - = Since P(C)=0.6, we know ( ) 0.4 C P C = , hence ( ) 0.4 0.04 0 0.2 0.16 C C P AB C = - - - = We know then P(ABC)+P(AB C C)=0.5=P(AC) so ( ) ( ) ( ) ( ) 0.7 0.6 0.5 0.8 P A C P A P C P A C = + - = + - = a. P(A C BC C ) = 0 b. We can number the 8 cells from left to right, from top to bottom as 1 to 8, then we know that C A C are the boxes 1,2,5,6,7,8, and A C are the boxes 1,2,3,4,5,6; hence the conditional probability is 0.5 0.05 0.05 0.75 0.5 0.05 0.05 0 0.2 + + = + + + + c. The probability is 0 since C A C has no common set with A C . d. Since C B A and A C certainly has no intersection, we can easily use the rule ( ) ( ) ( ) ( ) ( ) ( ) P X Y P X P Y
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/21/2008 for the course EE 351k taught by Professor Bard during the Fall '07 term at University of Texas at Austin.

Page1 / 3

Solution to HW 2 - Solution to EE351K HW assignment 2 1 ABC...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online