During the repair of a large number of car engines it was found that part number 100 was changed in 36% and part number 101 in 42% of the cases, and that both parts were changed in 30% of cases.Is the replacement of part 100 connected with that of part 101?
Find the probability that in repairing an engine for which part 100 has been changed it will also be necessary to replace part 101.
probability
2007-12-05 12:00 am
回答 (2)
2007-12-05 12:33 am
✔ 最佳答案
Assume that there are total of 100 car enginesBy assumtion, there are 36 engines repaired part 100 and 30 engines repaired both
Concerning only the ones repaired part 100, there are total of 36 possible outcomes
30 of them satisfy the requirement to replace part 101
Thus,
P(Replace part 101/part 100)
=30/36
=5/6
OR
By the first principle,
P(Replace part 101/part 100)
= P(Replace part 101 and part 100)/P(Replace part 100)
=30/36
=5/6
參考: 自己
2007-12-05 12:38 am
Is the replacement of part 100 connected with that of part 101?
Yes. Since P(#100 change) x P(#101 change) is not equal to P(#100 and #101 change).
36% x 42% = 15.12% < 30%, i.e they are not independent events.
Find the probability that in repairing an engine for which part 100 has been changed it will also be necessary to replace part 101.
It is a conditional probability.
P(#101 change/#100 change) = P(#100 and #101 change) / P(#100 change)
= 30% / 36%
= 83.3%
Yes. Since P(#100 change) x P(#101 change) is not equal to P(#100 and #101 change).
36% x 42% = 15.12% < 30%, i.e they are not independent events.
Find the probability that in repairing an engine for which part 100 has been changed it will also be necessary to replace part 101.
It is a conditional probability.
P(#101 change/#100 change) = P(#100 and #101 change) / P(#100 change)
= 30% / 36%
= 83.3%
參考: self
收錄日期: 2021-04-13 14:38:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071204000051KK01407