A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?
Isn’t that a trick question? It says only a white counter is put in this bag. And the “either black or white” makes you assume there’s a 100% chance of drawing a white counter when “a white counter is put in”.
50%, u took out the white counter that you put, now there is a counter which is either white or black
I can wait till all the slots are occupied - 0%, 25%, 33.33%, 66.66% on the block
Well it would be a 50/50 chance if they were mutually exclusive and you can only put one in… but refer to my first post? (ie. a bag contains “A” counter, known to be black or white… a “white” one is put in…"). I don’t know if I’m an idiot but it doesn’t seem to get simpler than that.
and very clever quip by dinesh.sundrani I must add… hope you sense the sarcasm.
This was the only one-in-a-lifetime opportunity to have caught me off cringing to my sofa, crying in all seriousness and cribbing about GRE. Ok one more hint - Bertrand’s Box Paradox … Go Figure!!
Alright, you win, I have no clue what’s going on.
100%
“A bag contains a counter, known to be either white or black.” does this mean 1 counter with a 50/50 chance of white or black? If so, it’s pretty clumsy. If not, there is not enough information to do the problem.
The Official Answer to this is 2/3 and I want to know how?
Yeah, this is Bertrand’s paradox. So at the start in the box, there is either a white counter or a black counter with 50% prob on each. Now we put in a white counter so we know that there are two possible states in the bag each with 50% probability: State 1: W, W State 2: W, B We draw a conter from the box an observe it is white. What is the prob of white on next draw? So now we draw Bayes theorem (or just the definition of conditional prob) from our own CFAI box. P(White on Second Draw|W on First Draw) = P(W2 | W1) (notation) = P(W2 and W1)/P(W1) So in the numerator, P(W2 and W1) is the probability that we are in state 1 = 50%. P(W1) = P(W1|State 1)*P(State 1) + P(W1|State 2)*P(State 2) = 1*0.5 + 0.5*0.5 = 0.5+0.25 = 0.75 So putting these together P(White on Second Draw|W on First Draw) = 0.5/0.75 = 2/3
Simply awesome Joey! Thanks for the solution… I read the wiki twice and never did I understand that Bertrand’s paradox and a way to go about solving this probability problem.
Isn’t this just like the whole “Lets Make A Deal” theory, that says you have better odds of switching the door you chose after they show you the goat behind one of the two doors that you didn’t chose?
JoeyDVivre Wrote: ------------------------------------------------------- > Yeah, this is Bertrand’s paradox. > > So at the start in the box, there is either a > white counter or a black counter with 50% prob on > each. Now we put in a white counter so we know > that there are two possible states in the bag each > with 50% probability: > > State 1: W, W > State 2: W, B > > We draw a conter from the box an observe it is > white. What is the prob of white on next draw? > So now we draw Bayes theorem (or just the > definition of conditional prob) from our own CFAI > box. > > P(White on Second Draw|W on First Draw) = P(W2 | > W1) (notation) > > = P(W2 and W1)/P(W1) > > So in the numerator, P(W2 and W1) is the > probability that we are in state 1 = 50%. > > P(W1) = P(W1|State 1)*P(State 1) + P(W1|State > 2)*P(State 2) > = 1*0.5 + 0.5*0.5 = 0.5+0.25 = 0.75 > > So putting these together > > P(White on Second Draw|W on First Draw) = 0.5/0.75 > = 2/3 Nice. Btw, it’s a common interview question. I got the correct answer intuitively, but when I was trying to write a coherent solution, kept forgetting about dividing by 0.75
That is one of the all time great probability brain teaser’s King Kong! (Not sure that it exactly relates to this one exactly though)