In the following, A' will be used in place of Ac for ease of typography.
(1) since A and A' are partitions of A, we can apply the law of total probability
P(B)=P(B|A)*P(A)+P(B|A')*P(A')
(2) Use Bayes Theorem
P(A|B)
=P(A∩B)/P(B) ... def of cond. prob
=P(B∩A)/P(B) ... commutativity
=[P(B|A)*P(A)]/P(B) ... def of cond. prob
(3) We need to use
P(A'∩B')=1-P(A∪B) ... complement
and
P(A∪B)=P(A)+P(B)-P(A∩B) ... prob. of union of sets
P(A'|B')
=P(A'∩B')/P(B') .. def of cond.prob
=(1-P(A∪B))/(1-P(B)) ... compl
=(1-P(B∪A))/(1-P(B)) ... commutativity
=[1-(P(A)+P(B)-P(A∩B)]/(1-P(B)) ... union of sets
=[1-(P(A)+P(B)-P(B∩A)]/(1-P(B)) ... commutativity
=[1-(P(A)+P(B)-P(B|A)*P(A)]/(1-P(B)) ... def. of cond. prob.
Please check and understand every line. Post if you have questions.
Let A,B be events in a sample space. you are given the following probabilities:
P(A)=0.8, P(B|A)=0.3, P(B|A^c)=0.8
Find:
1) P(B)
2) P(A|B)
3) P(A^c|B^c)
1 answer