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Unformatted text preview: 1 &¡¢£¢ ¤¥ &¦§¨©ª¨«¬®«¯¦°± ¡²³©ª´µ 3.12 Each simple event is equally likely, with probability 1/5. a. S: E 1 , E 2 , E 3, E 4 , E 5 P(S) = 5/5 =1 b. A: E 1 , E 3 P(A) = 2/5 c. B: E 1 , E 2 , E 4 , E 5 P(B) = 4/5 d. C: E 3, E 4 P(C) = 2/5 e. Α : E 2, E 4, E 5 P( Α ) = 3/5 f . Β : E 3 P( Β ) = 1/5 g . AB: E 1 P(AB) = 1/5 h. AC: E 3 P(AC) = 1/5 i. A & B: E 3 P(A & B) = 1/4 j. A & B: S P(A & B) = 1 k . A & C: E 1 , E 3, E 4 P(A & C) = 3/5 l. A & B: E 3 P(A & B) = 1/2 3.13 . From Exercise 3.12, P (A and B) = 1/5, P (A & B) = 1/4, and P (A) = 2/5. Since P (A and B) ¡ 0, A and B are not mutually exclusive. Since P (A & B) ¡ P (A), A and B are not independent. 3.18 Define the following events: L: company maintains a parental leave program S: company pays salary H: company pays health care It is given that P (L) = .27; P (S & L) = 1/3; P (H & L) = ¾ 2 a. P (L and S) = P (L)*P (S & L) = .27 ( 3 1 ) = .09 b. P (L and Η ) = P (L)*P ( Η & L) = .27 ( 4 1 ) = .0675 3.22. There are 35 brands from which to choose, each of which is categorized according to quality. a. P (E) = 35 2 b. P ( at least “good”) = P ( G or V or E ) = P (G) + P (V) + P (E) = 35 34 35 21 35 2 35 11 = + + Note: Since G, V, and E are simple events, they are mutually exclusive. Therefore, the Since G, V, and E are simple events, they are mutually exclusive....
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This note was uploaded on 10/02/2009 for the course PSTAT 5E taught by Professor Eduardomontoya during the Spring '08 term at UCSB.
 Spring '08
 EduardoMontoya

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