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Solution to HW 2

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

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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 P X Y P X P Y = + - = + to get P(A C B)+P(AC)=0.05+0+0.5=0.55 e. ( ) 1 ( ) c c c P A B C P A B C = - =0.8 2. a. The maximum value of ( ) P A B is 1, when A doesn't overlap with B b. The minimum value of ( ) P A B is 0.6, when A is completely covered by B c. The maximum value of

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

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