The probability that a bearing meets a specification is 0.92. Six bearings are picked at random. Calculate the probability that
(a)
all six meet the specification;
(b)
more than four meet the specification;
(c)
one or none meets specification; and
(d)
and exactly four meet the specification.
Probability
2013-09-22 7:25 pm
回答 (1)
2013-09-22 9:00 pm
✔ 最佳答案
(a) P(6meet) = 0.92^6
= 0.606355001
(b) P(5meet) + P(6meet)
= C(6,5)*(0.08)*(0.92)^5 + 0.92^6
= 0.922714132
(c) P(1meet) + P(0meet)
= C(6,1)*(0.92)*(0.08)^5 + 0.08^6
= 0.00001835008
(d) P(4meet)
= C(6,4)*(0.92)^4*(0.08)^2
= 0.068773724
2013-09-22 13:01:41 補充：
http://hk.knowledge.yahoo.com/question/question?qid=7013083100186
收錄日期: 2021-04-20 14:21:38
原文連結 [永久失效]:
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