close
標題:
發問:
http://vivianchung0124.googlepages.com/IMG_0935.jpg i don't know how to do question Q.27-29
最佳解答:
(28) For (a) (ii), its probability is 0.9510. Probability of (a) (iii) = 10 x 0.05 x 0.959. So probability of (a) (i) is 1 - 0.9510 - 10 x 0.05 x 0.959. The prob. that the batch will be accepted or rejected after first inspection = 0.9510 + 1 - 0.9510 - 10 x 0.05 x 0.959 = 0.6849, say P. In this case, 10 bulbs are inspected. Then, prob that 20 bulbs will be inspected = 1 - P = 0.3151 So overall speaking, the expected no. of bulbs inspected is: 10P + 20(1 - P) = 13.15 (29) For a box: P(no def) = 0.9724 P(1 def) = 24 x 0.9723 x 0.03 P(2 def) = 24C2 x 0.9722 x 0.032 So, for (a), the prob. is 1 - P(no def) = 0.5186 (b) (i) Prob = 0.51864 = 0.0723 (ii) Prob = [P(1 def)]4 = 0.0163 (c) (i) 100 x P(2 def) = 12.71 boxes (ii) P(At least 2 def) = 1 - P(no def) - P(1 def) = 0.1612 So expected no. of boxes = 16.12 (iii) P(Not more than 2 def) = P(No def) + P(1 def) + P(2 def) = 0.9659 So expected no. of boxes = 96.59 2008-05-25 22:38:27 補充： Q27 的答案太長, 我己圖像化如下: http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob1.jpg http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob2.jpg 2008-05-25 22:38:33 補充： http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob3.jpg http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob4.jpg
其他解答:
Q 27. Let Y be the number of questions he know the answer. Let Z be the number of questions guess correctly. P(X=3) = probability of having exactly 3 questions answer correctly = P(Z=3|Y=0) *P(Y=0) + P(Z=2|Y=1)*P(Y=1) + P(Z=1|Y=2)*P(Y=2) +P(Z=0|Y=3)*P(Y=3)
此文章來自奇摩知識+如有不便請留言告知
al applied maths probability發問:
http://vivianchung0124.googlepages.com/IMG_0935.jpg i don't know how to do question Q.27-29
最佳解答:
(28) For (a) (ii), its probability is 0.9510. Probability of (a) (iii) = 10 x 0.05 x 0.959. So probability of (a) (i) is 1 - 0.9510 - 10 x 0.05 x 0.959. The prob. that the batch will be accepted or rejected after first inspection = 0.9510 + 1 - 0.9510 - 10 x 0.05 x 0.959 = 0.6849, say P. In this case, 10 bulbs are inspected. Then, prob that 20 bulbs will be inspected = 1 - P = 0.3151 So overall speaking, the expected no. of bulbs inspected is: 10P + 20(1 - P) = 13.15 (29) For a box: P(no def) = 0.9724 P(1 def) = 24 x 0.9723 x 0.03 P(2 def) = 24C2 x 0.9722 x 0.032 So, for (a), the prob. is 1 - P(no def) = 0.5186 (b) (i) Prob = 0.51864 = 0.0723 (ii) Prob = [P(1 def)]4 = 0.0163 (c) (i) 100 x P(2 def) = 12.71 boxes (ii) P(At least 2 def) = 1 - P(no def) - P(1 def) = 0.1612 So expected no. of boxes = 16.12 (iii) P(Not more than 2 def) = P(No def) + P(1 def) + P(2 def) = 0.9659 So expected no. of boxes = 96.59 2008-05-25 22:38:27 補充： Q27 的答案太長, 我己圖像化如下: http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob1.jpg http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob2.jpg 2008-05-25 22:38:33 補充： http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob3.jpg http://i117.photobucket.com/albums/o61/billy_hywung/May08/Crazyprob4.jpg
其他解答:
Q 27. Let Y be the number of questions he know the answer. Let Z be the number of questions guess correctly. P(X=3) = probability of having exactly 3 questions answer correctly = P(Z=3|Y=0) *P(Y=0) + P(Z=2|Y=1)*P(Y=1) + P(Z=1|Y=2)*P(Y=2) +P(Z=0|Y=3)*P(Y=3)
文章標籤
全站熱搜
留言列表
