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P(A)=3//8,P(B)=1//2,P(AuuB)=5//8, which ...

`P(A)=3//8,P(B)=1//2,P(AuuB)=5//8,` which of the following do/does hold good?

A

`P(A^(C )//B)=2P(A//B^(C ))`

B

`P(B)=P(A//B)`

C

`15P(A^(c )//B^(C ))=8P(B//A^(C ))`

D

`P(A//B^(C ))=(AnnB)`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`(a,b,c,d)` `P(AuuB)=P(A)+P(B)-P(AnnB)`
`(5)/(8)=(3)/(8)+(1)/(2)-P(AnnB)`
`impliesP(AnnB)=(2)/(8)=(1)/(4)`
Now `P(A^(C )//B)=(P(A^(C )nnB))/(P(B))=(P(B)-P(AnnB))/(P(B))`
`=1-(1//4)/(1//2)=(1)/(2)`
`2P(A//B^(C ))=(2P(AnnB^(C )))/(P(B^(C )))`
`=(2(P(A)-P(AnnB)))/(1-P(B))`
`=4((3)/(8)-(2)/(8))=(1)/(2)=P(A^(C )//B)`
`P(A//B)=(P(AnnB))/(P(B))=(1)/(4)*(2)/(1)=(1)/(2)=P(B)`
Again `P(A^(C )//B^(C ))=(P(A^(C )nnB^(C)))/(P(B^(C )))=(1-P(AuuB))/(1-P(B))`
`=2(1-(5)/(8))=(3)/(4)`
`P(B//A^(C ))=(P(BnnA^(C )))/(1-P(A))=(P(B)-P(AnnB))/(5//8)`
`((1)/(2)-(1)/(4))/((5)/(8))=(1)/(4)*(8)/(5)=(2)/(5)`
Hence `8P(A^(C )//B^(C))=15P(B//A^(C ))`
`2P(A//B^(C ))=(1)/(2)`
`impliesP(A//B^(C ))=(1)/(4)=P(AnnB)`
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