Correct Answer - A::B::C::D
Let `D_(1)` denotes the occurrence of a defective bulb in Ist draw.
Therefore, `P(D_(1))=(50)/(100)=(1)/(2)`
and let `D_(2)` denotes the occurrence of a defective bulb in Iind draw.
Therefore, `P(D_(2))=(50)/(100)=(1)/(2)`
and let `N_(1)` denotes the occurrence of non-defective boulb in Ist draw.
Therefore, `P(N_(1))=(50)/(100)=(1)/(2)`
Again, let `N_(2)` denotes the occurrence of non-defective bulb in IInd draw.
Therefore, `P(N_(2))=(50)/(100)=(1)/(2)`
Now, `D_(1)` is independent with `N_(1)` and `D_(2)` is independent with `N_(2)`.
According to the given condition,
`A={"the first bulb is defective"}={D_(1)D_(2),D_(1)N_(2)}`
`B={"the second bulb is non-defective"}={D_(1)N_(2),N_(1)N_(2)}`
and `C={"the two bulbs are both defective"}={D_(1)D_(2),N_(1)N_(2)}`
Again, we know that,
`AnnB={D_(1)N_(2)},BnnC={N_(1)N_(2)}.`
`CnnA={D_(1)D_(2)}" and "AnnBnnC=phi`
Also, `P(A)=P{D_(1)D_(2)}+P{D_(1)N_(2)}`
`=P(D_(1))P(D_(2))+P(D_(1))P(N_(2))`
`=((1)/(2))((1)/(2))+((1)/(2))((1)/(2))=(1)/(2)`
Similarly, `P(B)=(1)/(2)" and "P(C)=(1)/(2)`
Also, `P(AnnB)=P(D_(1)N_(2))=P(D_(1))P(N_(2))=((1)/(2))((1)/(2))=(1)/(4)`
Similarly, `P(BnnC)=(1)/(4),P(CnnA)=(1)/(4)`
and `P(AnnBnnC)=0.`
Since, `P(AnnB)=P(A)P(B),P(BnnC)=P(B)P(C)`
and `" "P(CnnA)=P(C)P(A).`
Therefore, A, B and C are pairwise independent. Also, `P(AnnBnnC)neP(A)P(B)P(C)` therefore A,B and C cannot be independent.
