. a) If P(A)=0.6, P(B)=0.15 and P(B|A’)=0.25 find the following probabilities: i) P(B|A) [2 marks] ii) P(A|B) [2 marks] iii) p(AUB)?
回答 (2)
2020-06-20 12:26 am
i)
P(A) = 0.6
Hence, P(A') = 1 - P(A) = 0.4
P(B|A') = 0.25
P(B⋂A') / P(A') = 0.25
[P(B) - P(B⋂A)] / P(A') = 0.25
[0.15 - P(B⋂A)] / 0.4 = 0.25
0.15 - P(B⋂A) = 0.1
P(B⋂A) = 0.05
P(B|A)
= P(B⋂A) / P(A)
= 0.05 / 0.6
= 1/12
====
ii)
P(A|B)
= P(B⋂A) / P(A)
= 0.05 / 0.15
= 1/3
====
iii)
P(A⋃B)
= P(A) + P(B) - P(B⋂A)
= 0.6 + 0.15 - 0.05
= 0.7
P(A) = 0.6
Hence, P(A') = 1 - P(A) = 0.4
P(B|A') = 0.25
P(B⋂A') / P(A') = 0.25
[P(B) - P(B⋂A)] / P(A') = 0.25
[0.15 - P(B⋂A)] / 0.4 = 0.25
0.15 - P(B⋂A) = 0.1
P(B⋂A) = 0.05
P(B|A)
= P(B⋂A) / P(A)
= 0.05 / 0.6
= 1/12
====
ii)
P(A|B)
= P(B⋂A) / P(A)
= 0.05 / 0.15
= 1/3
====
iii)
P(A⋃B)
= P(A) + P(B) - P(B⋂A)
= 0.6 + 0.15 - 0.05
= 0.7
2020-06-20 12:09 am
P(B) = P(B|A) * P(A) + P(B|A') * P(A')
Or, 0.15 = P(B|A) * 0.6 + 0.25 * 0.4
Or, 0.15 = P(B | A) * 0.6 + 0.1
Or, P(B | A) = (0.15 - 0.1)/0.6
Or, P(B | A) = 0.083
P(A|B) = P(B|A) * P(A)/P(B)
= (0.083 * 0.6)/0.15 = 0.332
P(A U B) = P(A) + P(B) - P(A \cap B)
= 0.6 + 0.15 - (0.332 * 0.15)
= 0.6 + 0.15 - 0.0498 = 0.7002
Or, 0.15 = P(B|A) * 0.6 + 0.25 * 0.4
Or, 0.15 = P(B | A) * 0.6 + 0.1
Or, P(B | A) = (0.15 - 0.1)/0.6
Or, P(B | A) = 0.083
P(A|B) = P(B|A) * P(A)/P(B)
= (0.083 * 0.6)/0.15 = 0.332
P(A U B) = P(A) + P(B) - P(A \cap B)
= 0.6 + 0.15 - (0.332 * 0.15)
= 0.6 + 0.15 - 0.0498 = 0.7002
收錄日期: 2021-04-24 07:52:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200619155155AAKs9SP