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131A_1_midterm-2010-sol

# 131A_1_midterm-2010-sol - EE131A(Winter 2010 Mid-term...

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Mid-term solution 1. (a) P[A B]= P[A|B] P[B]=(0.5)(0.2)=0.1 P[A]= P[A B]-P[B]+P[A B]=0.8-0.2+0.1=0.7 (b ) P[A]P[B]=(0.7)(0.2)=0.14 P[A B]=0.1 So, they are not independent. Since P[A B] ≠ 0 , they are not mutually exclusive. (c ) P[ (A B) c ]=1-P[A B]=1-0.8=0.2 P[B c |A c ]=P[B c A c ]/P[A c ]=P[(A B) c ]/ P[A c ]=0.2/(1-0.7)=0.67 2. (a) P[Accident]=P[ Accident|| Good Risk s]P[ Good Risks ]+P [Accident | Average Risks ]P[ Average Risks ]+P [Accident | Bad Risk ]P[ Bad Risk ] = (0.05)(0.2)+(0.15)(0.5)+(0.3)(0.3)=0.175 (b) ( | ) ( | ) ( ) / ( ) (0.95)(0.2) /(1 0.175) 0.230 c c c P G A P A G P G P A = = - = 3. (a) P [World series is decided in 4 games] = P (Yankees w in the first four: WWWW ) + P (Yankees lose first four: LLLL ) = 2( 1 2 ) 4 = 1 8 (b) The outcomes where the Yankees win in 5 games are: Yankees w in exactly 3 of the first four gam es 1 2 4 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 4 Yankees w in gam e 5 1 2 4 4 4 3 4 4 4 Similarly, the outcomes where the Yankees lose in 5 games are: Yankees lose exactly 3 of the first four gam es 1 2 4 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 4 Yankees lose gam e 5 1 2 4 4 1 2 ) 5 = 1 4

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131A_1_midterm-2010-sol - EE131A(Winter 2010 Mid-term...

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