Proof that if two events are independent, so are their complements.

Just getting warmed up.
Here I prove that if events A and B are independent, so are Ac and Bc. I make use of De Morgan's Laws, without offering a formal proof of that part (but I do provide a brief Venn diagram justification of the needed bit).
More probability and statistics videos will follow.

[@rajatshukla2605]

Very very explained . Thanks for the video!

[@andyl.5998]

"More probability and statistics videos will follow." Hell yeah!

[@prabhleenreen3594]

thank u king just what I was looking for

[@sanchoysamriddharanchi5753]

welcome back.I am so happy.

[@SimchaWaldman]

This is beautiful! 😍🥰❤💚💙

[@sergiohenriqueandradedeaze9304]

Great channel!

[@elizaldodesousajunior9077]

Excellent video! 10 10 10!!!

[@AlexandreLopez1999]

Maybe worth noting that the backward implication also holds! So if the complements are independent, then so are the events themselves. If we let A = Xᶜ and B = Yᶜ be independent, then (by this vid) Aᶜ = Xᶜᶜ=X and Bᶜ=Yᶜᶜ = Y are also indpendent

[@haomiaoyu9813]

Thank you!

[@chizuru971]

Thank you 😇

[@nasser-eddinebendaoud6783]

Thanks Very Helpful

[@chikeconcepta2245]

keep them videos coming

[@marimkrtchyan2588]

Thanks!!!

[@HaineGratuite]

welcome back!!

[@jbstatistics]

Thanks!

[@lpm7656]

Can we generalize this result with n set ?

[@shaycohen622]

THX!

[@F1U7R2Y9]

Thumbnail looks Grant's video
But its close

[@andrealupascu295]

is it also true that if Ac and B are indepedent, A and Bc also are? (without knowing if A and B are indepedent)

[@jbstatistics]

Yes. If Ac and B are independent, so are A and B, A and Bc, and Ac and Bc. Independence of any one of those pairs implies independence of the others. Loosely, independence means that knowing that one event happened (or didn't happen) doesn't change the probability of the other event.

[@andrealupascu295]

@@jbstatistics thank you

Let A and B be independent events. Let C = A ∪ B. Conditional on C, are A and B independent?

[@dfghj241]

isn't that David Hume's?

[@jbstatistics]

I confess to not being well versed in the full historical context (and I'm sure the knowledge that "not (A or B)" and "not A and not B" are the same thing goes *way* back, long before De Morgan), but in probability that notion is typically referred to as one of De Morgan's Laws.

[@senselessnothing]

These were tough times to do probability.

[@rozebeef7427]

@wilma rumscheißen