Homework Solutions 01 - Solutions to Homework#1(W4150 Intro...

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Solutions to Homework #1 (W4150 ‘Intro to Probability and Statistics’, S04) Sec. 2.2. (1) (a) S = {8,16,24,32,40,48} (b) x 2 + 4x – 5 = (x + 5)(x – 1) = 0 x = -5 and x =1, i.e. S = {-5,1} (c) S = {T, HT, HHT, HHH} (d) S = {N. America, S. America, Europe, Asia, Africa, Australia, Antartica} (e) Solving 2x – 4 > 0 gives x > 2; but we must have also x < 1 S = Sec. 2.2. (4) (a) S = {(1,1), (1,2), …., (6,5), (6,6)} – set which consists of all possible pairs of (x,y) where both x and y can assume values of 1,2,3,4,5 and 6. The same could be expressed as a rule: (b) S = {(x,y)| 1 < x,y < 6} Sec. 2.2. (8) (a) A = {(x,y)| x + y > 8} = {(3,6), (4,5), (4,6), (5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)} (b) B = {(x,2) (2,y)| 1 < x,y < 6} = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)} (c) C = {(x,y)| 5 < x < 6, 1 < y < 6} = {(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} (d) A C = {(5,4), (5,5), (5,6), (6,3), (6,4), (6,5), (6,6)} (e) A B = (f) B C = {(5,2), (6,2)} (g) Sec. 2.2. (16) (a) M N = {x | 0 < x < 9} (b) M N = {x | 1 < x < 5} (c) M’ N’ = {x | 9 < x < 12}
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Sec. 2.3. (6) (a) By Th. 2.8 there are C 7 5 = 7!/(5!2!) = 21 ways (b) By the same theorem there are C 5 3 = 5!/(3!2!) = 10 ways Sec. 2.3. (9) With n 1 = 3 race cars, n 2 = 5 brands of gasoline, n 3 = 7 test sites and n 4 = 2 drivers, the generalized multiplication rule yields n 1 * n 2 * n 3 * n 4 = 210 test runs.
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