Question #207238

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


Expert's answer

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

A.


P(B∩A′)=P(B)−P(A∩B)P(B\cap A')=P(B)-P(A\cap B)

=0.6−0.35=0.25=0.6-0.35=0.25

B.


P(B′)=1−P(B)P(B')=1-P(B)

=1−0.6=0.4=1-0.6=0.4

C.


P(A)P(B)=0.7(0.6)=0.42≠0.35=P(A∩B)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(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)

=0.7+0.6−0.35=0.95=0.7+0.6-0.35=0.95

E.


P(B∣A)=P(A∩B)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(A∩B)=0.35≠0P(A\cap B)=0.35\not=0

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



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