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Unformatted text preview: 1 & 3.14. The eight simple events and their associated probabilities are given in the exercise. a. P (A) = P (E 1 ) + P (E 4 ) + P (E 6 ) = .1 + .05 + .2 = .35 b. P (B) = .05 + .05 + .3 + .2 + .1 = .70 c. P (AB) = P (E 4 ) + P (E 6 ) = .05 + .2 = .25 d. P (A & B) = P (B) + P (E 1 ) = .70 + .1 = .80 e. Since P (AB) & 0, A and B are not mutually exclusive. f. P (A B) = P (AB) / P (B) = .25/.70 = .3571 g. P (B A) = P (AB) / P (A) = .25/.35 = .7143 h. P (A B) & P (A), A and B are not independent. 3.16. Define the following events: A: new order exceeds $1000 B: new order exceeds $2000 C: new order exceeds $3000 D: new order is $2000 or less a. P (B) = .25 + .20 +.10 = .55 b. P (D A) = P (AD) / P (A) = .35/.90 = .39 c. P (C B) = P (BC) / P (B) = .30/.55 = .545 3.19. Define the following events: A: car has been in an accident in the past year B: car has antilock brakes The four cells of the probability table represent the intersection probabilities AB, A B , A B, and A B . a. P (A) = P (AB) + P (A B ) = .03 +.12 =.15 b. P ( A B) = .40 (directly from the table) c. P (B A) = P(A) P(AB) = .15 .03 = .2 3.20. a. The probability that the first car has antilock brakes is .43, since that is the proportion of cars in the large group of 1993model cars having antilock brakes. Whether or not the first car in the large group of 1993model cars having antilock brakes....
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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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