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Unformatted text preview: Chapter 1 Probability Theory “If any little problem comes your way, I shall be happy, if I can, to give you a hint or two as to its solution.” Sherlock Holmes The Adventure of the Three Students 1.1 a. Each sample point describes the result of the toss (H or T) for each of the four tosses. So, for example THTT denotes T on 1st, H on 2nd, T on 3rd and T on 4th. There are 2 4 = 16 such sample points. b. The number of damaged leaves is a nonnegative integer. So we might use S = { , 1 , 2 ,... } . c. We might observe fractions of an hour. So we might use S = { t : t ≥ } , that is, the half infinite interval [0 , ∞ ). d. Suppose we weigh the rats in ounces. The weight must be greater than zero so we might use S = (0 , ∞ ). If we know no 10dayold rat weighs more than 100 oz., we could use S = (0 , 100]. e. If n is the number of items in the shipment, then S = { /n, 1 /n,..., 1 } . 1.2 For each of these equalities, you must show containment in both directions. a. x ∈ A \ B ⇔ x ∈ A and x / ∈ B ⇔ x ∈ A and x / ∈ A ∩ B ⇔ x ∈ A \ ( A ∩ B ). Also, x ∈ A and x / ∈ B ⇔ x ∈ A and x ∈ B c ⇔ x ∈ A ∩ B c . b. Suppose x ∈ B . Then either x ∈ A or x ∈ A c . If x ∈ A , then x ∈ B ∩ A , and, hence x ∈ ( B ∩ A ) ∪ ( B ∩ A c ). Thus B ⊂ ( B ∩ A ) ∪ ( B ∩ A c ). Now suppose x ∈ ( B ∩ A ) ∪ ( B ∩ A c ). Then either x ∈ ( B ∩ A ) or x ∈ ( B ∩ A c ). If x ∈ ( B ∩ A ), then x ∈ B . If x ∈ ( B ∩ A c ), then x ∈ B . Thus ( B ∩ A ) ∪ ( B ∩ A c ) ⊂ B . Since the containment goes both ways, we have B = ( B ∩ A ) ∪ ( B ∩ A c ). (Note, a more straightforward argument for this part simply uses the Distributive Law to state that ( B ∩ A ) ∪ ( B ∩ A c ) = B ∩ ( A ∪ A c ) = B ∩ S = B. ) c. Similar to part a). d. From part b). A ∪ B = A ∪ [( B ∩ A ) ∪ ( B ∩ A c )] = A ∪ ( B ∩ A ) ∪ A ∪ ( B ∩ A c ) = A ∪ [ A ∪ ( B ∩ A c )] = A ∪ ( B ∩ A c ) . 1.3 a. x ∈ A ∪ B ⇔ x ∈ A or x ∈ B ⇔ x ∈ B ∪ A x ∈ A ∩ B ⇔ x ∈ A and x ∈ B ⇔ x ∈ B ∩ A . b. x ∈ A ∪ ( B ∪ C ) ⇔ x ∈ A or x ∈ B ∪ C ⇔ x ∈ A ∪ B or x ∈ C ⇔ x ∈ ( A ∪ B ) ∪ C . (It can similarly be shown that A ∪ ( B ∪ C ) = ( A ∪ C ) ∪ B .) x ∈ A ∩ ( B ∩ C ) ⇔ x ∈ A and x ∈ B and x ∈ C ⇔ x ∈ ( A ∩ B ) ∩ C . c. x ∈ ( A ∪ B ) c ⇔ x / ∈ A or x / ∈ B ⇔ x ∈ A c and x ∈ B c ⇔ x ∈ A c ∩ B c x ∈ ( A ∩ B ) c ⇔ x / ∈ A ∩ B ⇔ x / ∈ A and x / ∈ B ⇔ x ∈ A c or x ∈ B c ⇔ x ∈ A c ∪ B c . 1.4 a. “ A or B or both” is A ∪ B . From Theorem 1.2.9b we have P ( A ∪ B ) = P ( A )+ P ( B ) P ( A ∩ B ). 12 Solutions Manual for Statistical Inference b. “ A or B but not both” is ( A ∩ B c ) ∪ ( B ∩ A c ). Thus we have P (( A ∩ B c ) ∪ ( B ∩ A c )) = P ( A ∩ B c ) + P ( B ∩ A c ) (disjoint union) = [ P ( A ) P ( A ∩ B )] + [ P ( B ) P ( A ∩ B )]...
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 Spring '08
 Maceachern,S
 Probability, Probability theory, Trigraph, Sample points

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