I heard you betas were pretty smart. Can you help me with a homework problem?
>There are 3 boxes: One box has 2 white balls, one box has 2 black balls, and one box has one of each.
>Someone picks a box at random and pulls out a white ball
>What's the probability the other ball in that box is white?
Fifty percent. Oregano.
>I heard you betas were pretty smart
whoever told you this is a liar. i'm a hs dropout m8
just open everything, eventually there'll be a white box.
three fifths, it's a compromise
Math is racist
lt's two thirds.
Homework problems belong in So I'm not going to bother to tell you why the answer is 2/3
ooga booga where the complex surds at?
Thanks that makes sense. I guess you guys really are smart.
sounds like an iteration on the Monty hall problem, in which case the answer is absolutely not 50%. Not gonna tell you the real answer though. Do your own homework.
dropout indeed; the problem is about finding white balls, not white boxes lmfao
It's either in there.
Or it's not.
Nah, you're just dumb lol
Haha, yeah a little I guess. I get it's 50-50 now though
I'll give you a hint, OP. Both you and that faggot are wrong as everyone else who gets the stupid Monty Hall problem wrong decades later. You're gonna need Bayes' Theorem.
>Bayes' theorem
Yeah obviously. The probability given that the other ball is white given it is one of the two boxes containing a white ball is 1/2
I know how to show my work. Give me some credit
2/3. There's three white balls and two of them are partnered with another white ball.
that grill is cute
I don't get where 2/3 comes from.
If theres only 2 balls in the box, and it could only be black or white, where does the 3rd ball come from?
1/2 since you don't know which box it is and have to factor that in.
>inb4 some stupid technicality like 'muh 0/1 or 1/1'
fuck off, this is a realist thread.
They were just memeing to try to mess up OP's homework. It's clearly 1/2
So the original problem had 3 balls in 1 box?
It's easy.
X-probability 2nd ball is white
Y-prob 1st ball is white
P(x|y) =?
P(X|Y) =P(X^Y) /P(Y)
P(X^Y) =1/3 only one box out of the 3 has two white balls
P(Y) = 2/3 same as above
P(X|Y) =(1/3)/(2/3)=1/2
Glad to help user