# The expectations of i c i d and i s are the same

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above. The expectations of I C ,I D and I S are the same. Hence E ( X ) = 4 E ( I H ) = 4 1 - ( 39 5 ) ( 52 5 ) - ( 13 1 )( 39 4 ) ( 52 5 ) ! 1 . 468 . 4. [8 points] A hat contains two black and two white cards. Pick two cards and let X be the number of black cards. Find E ( X ) and SD ( X ). Solution: Note that P ( X = 0) = 2 4 × 1 3 = 1 6 , P ( X = 1) = 2 ± 2 4 × 2 3 ² = 2 3 , and P ( X = 2) = 2 4 × 1 3 = 1 6 . Therefore, E ( X ) = 0 × 1 6 + 1 × 2 3 + 2 × 1 6 = 1 , E ( X 2 ) = 0 × 1 6 + 1 × 2 3 + 4 × 1 6 = 4 3 , V ar ( X ) = E ( X 2 ) - ( E ( X )) 2 = 4 3 - 1 2 = 1 3 , and ﬁnally SD ( X ) = p V ar ( X ) = 1 3 . Page 3

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Math 431: Exam 1 5. [10 points] Ten people sit randomly around a circular table. Six are men and four are women. How many men do you expect are not sitting next to a woman? Solution: Let I n be the indicator that the person in the i th chair is a man and he has men sitting on both sides. Then, for each n ∈ { 1 ,..., 10 } E ( I n ) = P ( I n = 1) = 6 10 × 5 9 × 4 8 = 1 6 . The number of men not sitting next to a woman is X = I 1 ··· I 10 and E ( X ) = 10 E ( I 1 ) = 10 6 = 5 3 . 6. Let X and Y be independent random variables with E ( X ) = 4, SD ( X ) = 3, E ( Y ) = 6, and SD ( Y ) = 4. (a) [4 points] Find E ( X + Y ) and SD ( X + Y ). Solution: E ( X + Y ) = E ( X ) + E ( Y ) = 10. Because X,Y are independent V ar ( X + Y ) = V ar ( X ) + V ar ( Y ) = 3 2 + 4 2 = 25, so SD ( X + Y ) = 25 = 5. (b) [8 points] Find an upper bound for
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The expectations of I C I D and I S are the same Hence E X...

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