plugin-HW_01_Solutions

plugin-HW_01_Solutions - HW #1 Solutions (Fall 2011)...

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1 HW #1 Solutions (Fall 2011) Assigned problems: 2.5, 2.10, 2.18, 2.19, 2.23 2.5 (Confirmed) (a) Locations of W 1 Locations of W 2 Load at B Load at C M A Probability E 1 E 2 E 3 0 0 0 0.15 × 0.2=0.03 B 500 0 5,000 0.045 X C 0 500 10,000 0.075 X X B 200 0 2,000 0.05 X X B B 700 0 7,000 0.075 X X X B C 200 500 12,000 0.125 X C 0 200 4,000 0.12 X C B 500 200 9,000 0.18 X X C C 0 700 14,000 0.30 X or, sorted by M A : Locations of W 1 Locations of W 2 Load at B Load at C M A Probability E 1 E 2 E 3 0 0 0 0.03 B 200 0 2,000 0.05 X X C 0 200 4,000 0.12 X B 500 0 5,000 0.045 X B B 700 0 7,000 0.075 X X X C B 500 200 9,000 0.18 X X C 0 500 10,000 0.075 X X B C 200 500 12,000 0.125 X C C 0 700 14,000 0.30 X (b) E 1 and E 2 are not mutually exclusive because these two events can occur together, for example, when the weight W 2 is applied at C, M A is 10,000 ft-lb; hence both E 1 and E 2 will occur.
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2 (c) The probability of each possible value of M A is tabulated in the last column of the table above. (d) P( E 1 ) = P( M A > 5,000) = 0.075 + 0.075 + 0.125 + 0.18 + 0.30 = 0.755 P( E 2 ) = P(1,000 M A 12,000) = 0.045 + 0.075 + 0.05 + 0.075 + 0.12 + 0.18 = 0.545 P( E 3 ) = 0.05 + 0.075 = 0.125 P( E 1 E 2 ) = P(5,000 < M A 12,000) = 0.075 + 0.075 + 0.18 = 0.33 P( E 1 E 2 ) = 1 – 0.03 = 0.97 P( 2 E ) = 0.03 + 0.125 + 0.3 = 0.455 or P( 2 E ) = 1 – P( E 2 ) = 1 – 0.545 = 0.455
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plugin-HW_01_Solutions - HW #1 Solutions (Fall 2011)...

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