Help me please...............
please
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F5 probability 2 (急)
2011-03-10 5:16 am
回答 (1)
2011-03-10 5:50 am
✔ 最佳答案
23(a) P(both B and C malfunction)= P(B malfuctions) P(C malfuction)
= 0.3 * 0.4
= 0.12
(b) P(do not work properly)
= P(A mal or (both B and C mal)
= P(A mal) + P(both B and C mal) - P(both A, B and C mal)
= 0.2 + 0.12 - 0.024
= 0.296
P(device works properly)
= 1 - 0.296
= 0.704
收錄日期: 2021-04-26 14:05:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110309000051KK01185