close
標題:
Probability (Argent)
發問:
1: A had 0.87 problems per car and B had 1.53 problems per car. If you purchased a car from A, what is the probability that the car will have,a) zero problems?b) 2 or fewer problems?2: A state lottery is conducted in which 6 winning numbers are selected from a total of 54 numbers. What is the probability that... 顯示更多 1: A had 0.87 problems per car and B had 1.53 problems per car. If you purchased a car from A, what is the probability that the car will have, a) zero problems? b) 2 or fewer problems? 2: A state lottery is conducted in which 6 winning numbers are selected from a total of 54 numbers. What is the probability that if 6 numbers are radomly selected, a) all 6 numbers will be winning numbers? b) 5 numbers will be winning numbers? c) none of the numbers will be winning numbers?
最佳解答:
(1) (a)Suppose that the probability of no. of problems follows Poisson distribution: 圖片參考：http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png with λ = 0.87 for A. Then P(zero problems) = e-0.87 = 0.4190 (b) P(One problem) = 0.87 × e-0.87 /1! = 0.3645 P(Two problems) = 0.872 × e-0.87 /2! = 0.1586 P(Two or fewer problems) = 0.4190 + 0.3645 + 0.1586 = 0.9420 (2) (a) Using the binomial distribution, we have: Total possibilities of drawing 6 numbers from 54 numbers = 54C6 So P(all 6 win) = 1/(54C6) = 1/25827165 = 0.00000003872 (b) For 5 winning numbers drawn, total possibilities are 6C5 × 48C1 = 288 So P(5 win) = 288/25827165 = 0.00001115 (c) For none of numbers will be winning, total possibilities are 48C6 = 12271512 So P(None win) = 12271512/25827165 = 0.4751
其他解答:
Probability (Argent)
發問:
1: A had 0.87 problems per car and B had 1.53 problems per car. If you purchased a car from A, what is the probability that the car will have,a) zero problems?b) 2 or fewer problems?2: A state lottery is conducted in which 6 winning numbers are selected from a total of 54 numbers. What is the probability that... 顯示更多 1: A had 0.87 problems per car and B had 1.53 problems per car. If you purchased a car from A, what is the probability that the car will have, a) zero problems? b) 2 or fewer problems? 2: A state lottery is conducted in which 6 winning numbers are selected from a total of 54 numbers. What is the probability that if 6 numbers are radomly selected, a) all 6 numbers will be winning numbers? b) 5 numbers will be winning numbers? c) none of the numbers will be winning numbers?
最佳解答:
(1) (a)Suppose that the probability of no. of problems follows Poisson distribution: 圖片參考：http://upload.wikimedia.org/math/5/5/9/55978f02e2b22e9a93943595030ecf64.png with λ = 0.87 for A. Then P(zero problems) = e-0.87 = 0.4190 (b) P(One problem) = 0.87 × e-0.87 /1! = 0.3645 P(Two problems) = 0.872 × e-0.87 /2! = 0.1586 P(Two or fewer problems) = 0.4190 + 0.3645 + 0.1586 = 0.9420 (2) (a) Using the binomial distribution, we have: Total possibilities of drawing 6 numbers from 54 numbers = 54C6 So P(all 6 win) = 1/(54C6) = 1/25827165 = 0.00000003872 (b) For 5 winning numbers drawn, total possibilities are 6C5 × 48C1 = 288 So P(5 win) = 288/25827165 = 0.00001115 (c) For none of numbers will be winning, total possibilities are 48C6 = 12271512 So P(None win) = 12271512/25827165 = 0.4751
其他解答:
此文章來自奇摩知識+如有不便請留言告知
我都唔識,sorry
文章標籤
全站熱搜
留言列表
