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Unformatted text preview: == 1 2 9 10 2 1 1 .... 6 7 8 1 .... 7 8 9 10 ! 2 ! 8 ! 10 )! ( ! ! x x x x x x x x x x x r n r n 45 2. P(A ∪ B ) = P(A) + P(B) – P(A ∩ B) = 0.8 + 0.4 – 0.3 = 0.9. 3. { A } ={2, 4, 6}, { B} ={1, 2, 3} SO A = {1,3,5 } and A ∩ B = {1,3} 4. Let {E} denote {declare themselves computer science major at entry}, {G} denote event that a student {Graduates as a computer science major}. Given : P(E)=0.30 P(G)=0.20 and P(E ∩ G) = 0.10. We want P( G E ∩ ). Now P(E)= P(E ∩ G) +P( G E ∩ ). So 0.30 = 0.10 +P( G E ∩ ) . So P( G E ∩ ) = 0.300.10= 0.20. Or using the 2 x 2 table of probabilities and fact that P( A ) = 1 P(A) and P(A ∩ B )=P(A)  P(A ∩ B)we have P( G E ∩ ) = 0.2 . (bold =given) (not bold= calculated) UNDERLINED = question 4 calculation. {G} { G } Marginal probability E 0.1= = P(E ∩ G) 0.2= (0.3  0.1) =P(E ∩ G ) 0.3 = P(E) E 0.1 = ( 0.2  0.1 ) = P( E G) 0.6 =(0.7  0.1) = P( E G ) (1.0  0.3) = P( E ) Marginal Probability 0.2 = P(G) (1.0  0.2) = 0.8 =P( G ) 1.0 =P(S)...
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 Fall '03
 Mendell
 Computer Science, 10%, 20%, 30%

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