Question #131455
If p1=p(A),p2=p(B),p3=P(A ∩ B), (p1,p2,p3>0) express the following in terms of p1,p2,p3

1)P(A ∪ B)'
2)P(A'∪ B')
3)P(A' ∩ B)
4)P(A' ∪ B)
5)P(A' ∩ B')
6)P(A ∩ B')
7)P(A | B)
8)P(B | A')
9)P[A' ∩ (A ∪ B)]
1
Expert's answer
2020-09-07T17:24:24-0400

1)

P(AB)=1P(AB)=1[P(A)+P(B)P(AB)]P(A ∪ B)' = 1-P(A \cup B) = 1 - [P(A)+P(B)-P(A \cap B)]

P(AB)=1(p1+p2p3)=1p1p2+p3\therefore P(A ∪ B)' = 1-(p1+p2-p3) = 1-p1-p2+p3


2)

P(AB)=1P(AB)=1p3P(A'∪ B')=1-P(A\cap B) = 1-p3


3)

P(AB)=P(B)P(AB)=p2p3P(A' \cap B)=P(B)-P(A\cap B)=p2-p3


4)

P(AB)=1P(AB)=1[P(A)P(AB)]P(A' ∪ B)=1-P(A\cap B')=1-[P(A)-P(A\cap B)]

P(AB)=1(p1p3)=1p1+p3P(A' ∪ B)=1-(p1-p3)=1-p1+p3


5)

P(AB)=1P(AB)=1[P(A)+P(B)P(AB)]P(A' \cap B') = 1-P(A \cup B) = 1 - [P(A)+P(B)-P(A \cap B)]

P(AB)=1(p1+p2p3)=1p1p2+p3\therefore P(A' \cap B') = 1-(p1+p2-p3) = 1-p1-p2+p3

6)

P(AB)=P(A)P(AB)=p1p3P(A \cap B')=P(A)-P(A\cap B)=p1-p3


7)

P(AB)=P(AB)P(B)=p3p2P(A | B)= \frac {P(A\cap B)}{P(B)}=\frac {p3}{p2}


8)

P(BA)=P(BA)P(A)=P(B)P(AB)1P(A)=p2p31p1P(B | A')=\frac {P(B\cap A')}{P(A')}=\frac {P(B)-P(A\cap B)}{1-P(A)}= \frac {p2-p3}{1-p1}


9)

P[A(AB)]=P(BA)=P(B)P(AB)P[A' ∩ (A ∪ B)]=P(B\cap A')=P(B)-P(A\cap B)

P[A(AB)]=p2p3P[A' ∩ (A ∪ B)]=p2-p3


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