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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 1993-model cars having antilock brakes. Whether or not the first car in the large group of 1993-model 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