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Unformatted text preview:  sand)P(sand) + P(2  clay)P(clay)+ P(2  rock)P(rock)] – P(2  clay)P(clay) = (100/250) + [(10/100)(100/250) + (2/100)(100/250) + (2/50)(50/250)] – (2/100)(100/250) = 112/250 = 0.448 (e) Dependent , because, for example, P(rock AND 0) = 0.176 [see (c)] ≠ P(rock)P(0) = (50/250)(212/250) = 0.170 7. (a) P(steep AND 3) = 2/80 = 1/40 (b) P(mild OR 2) = P(mild) + P(2) – P(mild and 2) = 30/80 + 10/80 – 4/80 = 36/80 = 9/20 (c) P(mild  1) = P(mild and 1)/P(1) = (10/80)/(21/80) = 10/21 (d) If independent, P (flat and 2) = P(flat) x P(2) = (40/80) x (10/80) = 400/6400 = 1/16 Therefore, the number of hills with flat and 2 is 80 x 1/16 = 5 8. 9. Variability in horsepower is clearly greater in the US and least in Sweden and Italy....
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 Spring '08
 Stolzenbach
 Probability, Boelter Hall, Stolzenbach 5732J Boelter, Boelter Hall Prof., Engineering Department CEE, LOS ANGELES Civil

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