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- P(A)+P(B)−2P(A∩B)
- P(A∩B)+2P(¯A∩B)
- P(A∪B)−P(A∩B)
- P(¯A)+P(¯B)−2P(¯A∩¯B)

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Solution

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A and B is represented by A∩¯¯¯¯B+¯¯¯¯A∩B

hence P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)={P(A)−P(A∩B)}+{P(B)−P(A∩B)} =P(A)+P(B)−2P(A∩B)

which is option (A).

⇒P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=[P(A)+P(B)−P(A∩B)]−P(A∩B)=P(A∪B)−P(A∩B)

⇒P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=[P(A)+P(B)−P(A∩B)]−P(A∩B)=P(A∪B)−P(A∩B)

which is option (C).

P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A)+1−P(¯¯¯¯B)−2{1−P(¯¯¯¯A∪¯¯¯¯B)}=2P(¯¯¯¯A∪¯¯¯¯B)−P(¯¯¯¯A)−P(¯¯¯¯B)={2P(¯¯¯¯A)+2P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)}−P(¯¯¯¯A)−P(¯¯¯¯B)=P(¯¯¯¯A)+P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)

P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A)+1−P(¯¯¯¯B)−2{1−P(¯¯¯¯A∪¯¯¯¯B)}=2P(¯¯¯¯A∪¯¯¯¯B)−P(¯¯¯¯A)−P(¯¯¯¯B)={2P(¯¯¯¯A)+2P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)}−P(¯¯¯¯A)−P(¯¯¯¯B)=P(¯¯¯¯A)+P(¯¯¯¯B)−2P(¯¯¯¯A∩¯¯¯¯B)

which is option (D).

and P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A∪B)+P(¯¯¯¯A∩B)∵¯¯¯¯¯¯¯¯¯¯¯¯¯A∩¯¯¯¯B=¯¯¯¯A∪B

and P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)=1−P(¯¯¯¯A∪B)+P(¯¯¯¯A∩B)∵¯¯¯¯¯¯¯¯¯¯¯¯¯A∩¯¯¯¯B=¯¯¯¯A∪B

by demorgan's law P(A∩¯¯¯¯B)=1−P(¯¯¯¯A∪B)⇒P(A∩¯¯¯¯B)+P(¯¯¯¯A∩B)={1−P(¯¯¯¯A)−P(B)+P(¯¯¯¯A∩B)}+P(¯¯¯¯A∩B)=P(A)−P(B)+2P(¯¯¯¯A∩B)

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