Hey guys where do i find some kewl hardish HS level probabilities exercises? Many thanks xd
Hey guys where do i find some kewl hardish HS level probabilities exercises? Many thanks xd
Hudson Parker
Other urls found in this thread:
en.wikipedia.org
en.wikipedia.org
twitter.com
Eli Hill
Come on guys i need to practice :((
Adrian Miller
About conditional probabilities?
Any specific subject?
I could give you some horrendous measure/probability theory exercises.
Andrew White
Anyway, the Monty Hall problem (as seen in another thread) is fun. Hint: write it all out.
>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Caleb Perez
I do fairly well on the pre probabilities like permutations and combinatorics, my worse is at proving with sets i guess, what i had in mind was more something to review set rules but yeah sure my dude
Evan Johnson
Sets. Nice. Okay, two simple ones first:
1) If [math]A\subseteq B[/math], show that If [math]\mathbb{P}(A)\leq\mathbb{P}(B)[/math].
2) Show that [math]\mathbb{P}(A\cup B)-\mathbb{P}(A\cap B)=\mathbb{P}(A)+\mathbb{P}(B)[/math].
If your current knowledge doesn't get you any further, use the definition of measure:en.wikipedia.org
It is a probability measure if [math]\mathbb{P}(\text{space set})=1[/math], but you don't need that here.
Connor Turner
2) Restated because of glitches:
2) Show that [math]\mathbb{P}(A\cup B)-\mathbb{P}(A\cap B)=\mathbb{P}(A)+\mathbb{P}(B)[/math].
If your current knowledge doesn't get you any further, use the definition of measure:en.wikipedia.org
It is a probability measure if
[math]\mathbb{P}(\text{space set})=1[/math],
but you don't need that here.
Jordan Carter
2) Again:
[math]\mathbb{P}(A\cup B)-\mathbb{P}(A\cap B)=\mathbb{P}(A)+\mathbb{P}(B)[/math].
If your current knowledge doesn't get you any further, use the definition of measure:
en.wikipedia.org
It is a probability measure if
[math]\mathbb{P}(\text{space set})=1[/math]
but you don't need that here.
Brayden Hill
Fuck it:
[math]\mathbb{P}(A\cup B)-\mathbb{P}(A\cap B)=\mathbb{P}(A)+\mathbb{P}(B)[/math]
Logan Ortiz
REEEEEEEEEEEEEE
[math] \mathbb{P}(A\cup B) - \mathbb{P}(A\cap B) = \mathbb{P}(A) + \mathbb{P}(B) [/math]
Justin Collins
This one looks simple by just drawing circles A and B and proving it directly but ill try to get everything written, starting to work on the first problem in 15 secs looks like its 1/2 for both after a first read, going to check if if im messing something on the (car) | (not door C) thing
Carter Bennett
Now if you want a terror exercise and know what a proof by induction is, use 2) and induction to prove the inclusion-exclusion principle: see picture.
Juan Lewis
Post a pic of your proof or look the answer up on math stack or something if you want your answer checked.
[math]f [/math] is just the probability measure [math]\mathbb{P} [/math] here.
Lincoln Russell
So on the monty hall problem i got 1/2 on either door A or B (duh) and after checking on wikipedia and it saying 2/3 of winning if you switch door im skeptical... i mean you have 3 doors, one has a goat, so its like you have 2 doors, if you chose the goat door (which is 1/2 because you know its not door 3) and keep you get goat if you switch you get car, if you chose car door (again 1/2 because 2 reasonable doors) and keep you get car, if you switch you get goat...
Going for the first proof, expect nothing good as i get them intuitively but cant explain why, also highly doubting i do the induction one, it was hard a year ago and its going to be hard now
Angel Sanders
oh and im assuming that
>A⊆B
is 'A is inside or equal to B'
Nicholas Mitchell
Yeah it goes right against your intuïtion, but the condition is set up such that your luck increases (by opening the wrong doors). Just pick door 1 (without loss of generality, because there would be no difference if you chose 2 or 3) and write out all posibilities.
You can find the answer here in a nice table:
en.wikipedia.org
Yes, both [math]A\subset B [/math] and [math]A\subseteq B [/math] mean the same (inside or equal).
To not be equal you would say proper subset: [math]A\subsetneq B [/math]
Jackson Murphy
This is what i got for the first proof, yeah its literally baby tier proving something but best i could think of
Luis King
You can't say the #A
Jacob Miller
Yeah that one was pretty shabby, this one looks irrefutable tho, also can i just write whatever i can take out of drawing the circles?
Jeremiah Martinez
This proof is correct yes. I also see that my exercise was written incorrectly.
To answer your question, it should always be a "general" drawing which is not always possible (your drawing is the general case though), so try to stick with the countable additivity definition. But yeah, drawings definitely help your intuition, so use it to your advantage (as long as it agrees with the countable additivity definition)
Austin Wilson
...
James Gutierrez
Dont worry about incorrect exercises, i have though that my teacher had put a complete problem as a trick question and when it was to prove something i proved that it wasnt true (it actually was).
On this one i had to think harder, not sure if i used your measure definition because wikipedia is really abstract and i might know about it but in a more basic way.
Also had to assume that P(A)
Jackson Peterson
Well your proof is basically correct because of equivalencies, but you go top down instead of bottom up.
Note that [math](B\cap \bar{A}) \subset \Omega [/math] and not an element.
The regular way of proving is [math] \mathbb{P}(B)=\mathbb{P}(A\cup (B\cap \bar{A}))=\mathbb{P}(A)+\mathbb{P}(B\cap \bar{A})\geq \mathbb{P}(A) [/math], because the latter one is non-negative.
Landon Fisher
These two being equal is where you used the definition, because the sets are disjoint.