al applied maths probability

(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

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)