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hw2_418_10

# hw2_418_10 - 1 When you draw the first marble you have...

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Unformatted text preview: 1. When you draw the first marble, you have three choices, red, blue and green. a) For replacement case, you still have the same number of choices when you draw the second marble. So, the sample space is S = { ( r,r ) , ( r,g ) , ( r,b ) , ( g,r ) , ( g,g ) , ( g,b ) , ( b,r ) , ( b,g ) , ( b,b ) } . b) For without replacement case, you cannot repeat the same color as the first one. So, the sample space is S = { ( r,g ) , ( r,b ) , ( g,r ) , ( g,b ) , ( b,r ) , ( b,g ) } . 2. E 1 means 6 appears on the first roll. So E 1 only contains { 6 } . E 2 means 6 appears on the second roll, and E 2 contains { x 1 , 6 } , x 1 = 1 , 2 , 3 , 4 , 5. Similarly, E n contains { x 1 ,x 2 ··· x n- 1 , 6 } , where x i = 1 , 2 , 3 , 4 , 5 for any i from 1 to n- 1. ( S ∞ n =1 E n ) means that 6 can be finally got at least one time if you roll infinitely many times. So, ( S ∞ n =1 E n ) C means that 6 never appears. 5. Let x i = 1 denotes the ith components working, and x i = 0 denotes the ith components failed. a) When each x i takes the value of either 0 or 1, with i from 1 to 5, then there are altogether 2 5 outcomes in the experiment. b)Considering all the unions of 1 and 2 working, or 3 and 4 working, or 1,3, and 5 working, we get W= { (1,1,0,0,0), (1,1,0,0,1),(1,1,0,1,0),(1,1,0,1,1),(1,1,1,0,0),(1,1,1,0,1), (1,1,1,1,0),(1,1,1,1,1), (0,0,1,1,0),(0,0,1,1,1),(1,0,1,1,0),(1,0,1,1,1), (0,1,1,1,0),(0,1,1,1,1),(1,0,1,0,1) } c) If 4 are 5 are failed, then we only consider all possible choices for the first three components. It is 2three components....
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hw2_418_10 - 1 When you draw the first marble you have...

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