Introductory Probability (2 Viewers)

[InteGrand]

For the inductive step, write it as (A1 ∩ … ∩ An) ∩ A_(n+1).

Then apply the two set result to this (can assume using strong induction). Then remember Pr((A1 ∩ … ∩ An)^c) = 1 – Pr(A1 ∩ … ∩ An). Then use the n-set result (inductive hypothesis).

 

[leehuan]

Wait hang on, just got around to looking at it again more closely

(A₁ ∩ … ∩ A_n) ∩ A_n+1 is all good but sorry which two set result am I meant to apply again?

 

[InteGrand]

Wait hang on, just got around to looking at it again more closely

(A₁ ∩ … ∩ A_n) ∩ A_n+1 is all good but sorry which two set result am I meant to apply again?

 

[leehuan]

Ohh right! Sorry excuse my lack of experience with strong induction lol

ps kawaiipotato is a huge stalker

 

[kawaiipotato]

Ohh right! Sorry excuse my lack of experience with strong induction lol

ps kawaiipotato is a huge stalker

What how did I just randomly get mentioned

 

[leehuan]

What how did I just randomly get mentioned

Got given rep with an annotation to if kawaiipotato was a stalker but because the system isn't allowing me to give rep back yet I'm like ok ceebs pm-ing I'll just insert it at the bottom of my post.
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Is it actually common to write for probability?

 

[kawaiipotato]

i am innocent

 

[leehuan]

Quick question: Are any of the following true statements (or true given specific conditions)

 

[Drongoski]

Quick question: Are any of the following true statements (or true given specific conditions)

Both False


For 1st, can use a counter example. Throw a die. Consider the 2 events:

A = more than 2 = {3, 4,5, 6} and B = even = {2, 4, 6} and B'= {1, 3, 5}

P(A) = 4/6 = 2/3

P(A|B) = P({2, 3})/P(2,4,6}) = 2/3

P(A|B') = P(3,5)/P(1, 3, 5) = 2/3

.: P(A|B) + P(A|B') = 2/3 + 2/3 > 1 (an impossible event) =/= P(A)


For 2nd:

Given B (or for that matter, any other event), the prob of getting A or not-A is 1, regardless of what happens. But using the formulae:

P(A|B) + P(A'|B) = P(AB)/P(B) + P(A'B)/P(B) = (P(AB) + P(A'B))/P(B) = P(B)/P(B) = 1



But don't take my word for it; I may well be wrong. Did all these ages ago

 

[InteGrand]

Quick question: Are any of the following true statements (or true given specific conditions)

Those are both false as Drongoski has said (counter-examples may be found). A correction would be to use intersections rather than conditioning:

Pr(A) = Pr(A∩B) + Pr(A∩B^c).

Using the rule Pr(A∩B) = Pr(B)Pr(A|B), we can write this as

Pr(A) = Pr(A|B)Pr(B) + Pr(A|B^c)Pr(B^c), which is basically what you were after I think.

(Basically the Law of Total Probability)

 

[leehuan]

Ok yep law of total probability was what left my brain lol. Thanks

 

[leehuan]

Suppose you didn't know what the formula for conditional probability was. Would there be a way to tackle this question?

 

[InteGrand]

Suppose you didn't know what the formula for conditional probability was. Would there be a way to tackle this question?

This is just showing that if P(X|Y) = P(X), then X and Y are independent.

By definition, P(X|Y) = P(X,Y)/P(Y). (, meaning intersection).

Since P(X|Y) = P(X) (by assumption), we have

P(X,Y)/P(Y) = P(X) ==> P(X,Y) = P(X)P(Y).

Also, P(Y|X) = P(Y,X)/P(X) = P(X,Y)/P(X) = P(X)P(Y)/P(X) = P(Y).

(Assuming X and Y are events with positive probability.)

In order to do this Q., we would need to know the definition of conditional probability, because otherwise P(X|Y) wouldn't make sense to us.

Btw, we can also show that if X and Y are independent, then so are A and B, where A is any member of {X, X^c} and B is any member of {Y, Y^c}.

 

[leehuan]

This is just showing that if P(X|Y) = P(X), then X and Y are independent.

By definition, P(X|Y) = P(X,Y)/P(Y). (, meaning intersection).

Since P(X|Y) = P(X) (by assumption), we have

P(X,Y)/P(Y) = P(X) ==> P(X,Y) = P(X)P(Y).

Also, P(Y|X) = P(Y,X)/P(X) = P(X,Y)/P(X) = P(X)P(Y)/P(X) = P(Y).

(Assuming X and Y are events with positive probability.)

Yeah lol that's what I did; it's true by just the formula (which I feel is just a defintion)

But cause they had a, b, c, d I was like is there a way to bypass the formula


Edit: I should probably mention that this isn't uni homework or something for my own sake. Apparently this 2U class were taught set theory or something...

 

[InteGrand]

Yeah lol that's what I did; it's true by just the formula (which I feel is just a defintion)

But cause they had a, b, c, d I was like is there a way to bypass the formula

Well yeah, they probably want you to say things like P(X|Y) = b/(b+c) (via their diagram). But this just uses the definition anyway and makes the proof more tedious than it needs to be).

 

[leehuan]

Well yeah, they probably want you to say things like P(X|Y) = b/(b+c) (via their diagram). But this just uses the definition anyway and makes the proof more tedious than it needs to be).

Exactly what I reckoned.

 

[InteGrand]

Yeah lol that's what I did; it's true by just the formula (which I feel is just a defintion)

But cause they had a, b, c, d I was like is there a way to bypass the formula


Edit: I should probably mention that this isn't uni homework or something for my own sake. Apparently this 2U class were taught set theory or something...

Ah it's a HSC level Q. I was just about to say that if this was in HSC, they'd probably want you to use the b/(b+c) stuff. But then I remembered that conditional probability isn't taught (explicitly) in the HSC. If this 2U class was taught it, surely they'd have learnt the definition of conditional probability, because otherwise it wouldn't make sense to ask Q's about it. Anyway, this shouldn't be examinable in HSC (unless it's a topic in the new syllabus).