My proof therefore far:$$egingatheredPleft(A^Ccap B^C ight)=left(1-Pleft(A ight) ight)left(1-Pleft(B ight) ight)=\1-Pleft(B ight)-Pleft(A ight)+Pleft(A ight)Pleft(B ight)=1-Pleft(B ight)-Pleft(A ight)+Pleft(Acap B ight)endgathered$$

After that, I"m stuck. Any help would be appreciated.

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Assume $A$ and also $B$ room independent. TheneginalignP(A^c cap B^c) &= 1 - P(A cup B) \&= 1 - P(A) - P(B) + P(A cap B) \&= 1 - P(A) - P(B) + P(A)P(B) \&= (1-P(A))(1-P(B)) \&= P(A^c)P(B^c).endalign

gradient23"s proof is great, in my opinion, but I would favor to show an additional proof that seems an ext intuitive to me, though much less rigorous.

The evidence is based upon a verbal definition of freedom from wikipedia:

two occasions are elevation <...> if the occurrence of one go not impact the probability of incident of the other

In addition, we usage the fact that freedom is symmetric.

The (non-rigorous) proof:

We assume that $A$ and $B$ space independent.By definition, the occurrence of $A$ doesn"t influence the probability of $B$.Thus, the incident of $A$ also doesn"t affect the probability the $B^C$.So by definition, $A$ and also $B^C$ are likewise independent, i beg your pardon by an interpretation again way that the incident of $B^C$ doesn"t impact the probability the $A$.(Here we offered the the contrary of independence.)Therefore, the event of $B^C$ likewise doesn"t impact the probability the $A^C$.So by definition, $B^C$ and $A^C$ are likewise independent.See more: Nicki Minaj See Through Shirt, Nicki Minaj Goes Braless, Wears See

One could transform the proof to the language that carolannpeacock.com:$$egingatheredA ext and also B, extare independent\downarrow\Pleft(B|A ight)=Pleft(B ight)\downarrow\1-Pleft(B^C|A ight)=1-Pleft(B^C ight)\downarrow\Pleft(B^C|A ight)=Pleft(B^C ight)\downarrow\A ext and also B^C, extare independent\downarrow\Pleft(A|B^C ight)=Pleft(A ight)\downarrow\1-Pleft(A^C|B^C ight)=1-Pleft(A^C ight)\downarrow\Pleft(A^C|B^C ight)=Pleft(A^C ight)\downarrow\B^C ext and A^C, extare independentendgathered$$

But currently we supplied conditional probabilities, which can be a problem in instance $P(A)=0$ or $P(B^C)=0$ (here is a discussion about this problem).