Question #207238

Given that P(A)=0.7,P(B)= and P (A and B)=0.35


1
Expert's answer
2021-06-16T05:35:18-0400

Given P(A)=0.7,P(B)=0.6,P(A)=0.7, P(B)=0.6, and P(AB)=0.35.P(A\cap B)=0.35.

A.


P(BA)=P(B)P(AB)P(B\cap A')=P(B)-P(A\cap B)

=0.60.35=0.25=0.6-0.35=0.25

B.


P(B)=1P(B)P(B')=1-P(B)

=10.6=0.4=1-0.6=0.4

C.


P(A)P(B)=0.7(0.6)=0.420.35=P(AB)P(A)P(B)=0.7(0.6)=0.42\not=0.35=P(A\cap B)

Therefore AA and BB are dependent events.


D.


P(AB)=P(A)+P(B)P(AB)P(A\cup B)=P(A)+P(B)-P(A\cap B)

=0.7+0.60.35=0.95=0.7+0.6-0.35=0.95

E.


P(BA)=P(AB)P(A)P(B|A)=\dfrac{P(A\cap B)}{P(A)}

=0.350.7=0.5=\dfrac{0.35}{0.7}=0.5

F.


P(AB)=0.350P(A\cap B)=0.35\not=0

Therefore AA and BB are not mutually exclusive dependent events. e



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