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CEE 110 PS2 11 Solutions

# CEE 110 PS2 11 Solutions - f P(A’ ∪ B =(70 9 5/100 =...

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UNIVERSITY OF CALIFORNIA, LOS ANGELES Civil and Environmental Engineering Department CEE 110 Introduction to Probability and Statistics for Engineers Spring Quarter 2011 M-W 2-4 PM Franz 1178 Prof. K. D. Stolzenbach 5732J Boelter Hall, 206-7624 Problem Set 2 Solutions ___________________________________________________________________________ 1. a) { ab, ac, ad, bc, bd, cd, ba, ca, da, cb, db, dc } b) { ab, ac, ad, ae, af, ag, bc, bd, be, bf, bg, cd, ce, cf, cg, ef, eg, fg, ba, ca, da, ea, fa, ga, cb, db, eb, fb, gb, dc, ec, fc, gc, fe, ge, gf, de, df, dg, ed, fd, gd } c) Let d and g denote defective and good, respectively. Then S = { gg, gd, dg, dd } d) S = { gd, dg, gg } 2. All outcomes are equally likely a) P(A) = 2/5 b) P(B) = 3/5 c) P(A') = 3/5 d) P(A B) = 1 e) P(A B) = P( )= 0 3. a) P(A) = 86/100 = 0.86 b) P(B) = 79/100 = 0.79 c) P(A') = 14/100 = 0.14 d) P(A B) = 70/100 = 0.70 e) P(A B) = (70+9+16)/100 = 0.95

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Unformatted text preview: f) P(A’ ∪ B) = (70+9+5)/100 = 0.84 4. a) P(A') = 1- P(A) = 0.7 b) P ( ) = P(A) + P(B) - P( ) = 0.3+0.2 - 0.1 = 0.4 c) P( ) + P( ) = P(B). Therefore, P( ) = 0.2 - 0.1 = 0.1 d) P(A) = P( ) + P( ). Therefore, P( ) = 0.3 - 0.1 = 0.2 e) P(( )') = 1 - P( ) = 1 - 0.4 = 0.6 f) P( ) = P(A') + P(B) - P( ) = 0.7 + 0.2 - 0.1 = 0.8 5. a ) 0.82 b) 0.90 c) 8/9 = 0.889 d) 80/82 = 0.9756 e) 80/82 = 0.9756 f) 2/10 = 0.20 6. Let A denote the event that a respondent is a college graduate and let B denote the event that an individual votes for Bush. P(B) = P(A)P(B|A) + P(A’)P(B|A’) = 0.38 0.53+0.62 0.5 = 0.5114 7. P( ) = 70/100, P(A) = 86/100, P(B) = 77/100 Then, P( ) P(A)P(B), so A and B are not independent. 8. Let A denote the event that a respondent is a college graduate and let B denote the event that a voter votes for Bush. 9. Variability in horsepower is clearly greater in the US and least in Sweden and Italy....
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CEE 110 PS2 11 Solutions - f P(A’ ∪ B =(70 9 5/100 =...

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