統計學機率問題 機率分配..

(a) P(A or B) = P(A) + P(B) - P(AB)= 0.3 + 0.6 - 0.18= 0.72(b) P(A' or B')= P(A') + P(B') - P(A'B')= 0.7 + 0.4 - (1 - 0.72)= 1.1 - 0.28= 0.82(c) P(A'B') = 1 - P(A or B) = 0.28(d) 不是﹐因為P(AB) != 0 (e) P(AB) = 0.18, P(A)P(B) = 0.18 因此A,B獨立

(a)P(A∪B)
=P(A)+P(B)-P(A∩B)
=0.3+0.6-0.18
=0.72

(b)P(A'∪B')
=P(A')+P(B')-P(A'∩B')
=1-P(A)+1-P(B)-[1-P(A∪B)]
=1-0.3+1-0.6-1+0.72
=0.82

(c)P(A'∩B')
=1-P(A∪B)
=1-0.72
=0.28

(d)∵P(A∩B)≠0
∴A、B 不互斥

(e)∵P(A∩B)=P(A)P(B)
∴A、B 為獨立