More marbles
You have two identical looking bags. One bag has three black marbles and one white marble. The other has three white marbles and one black marble. After choosing a bag at random, you draw one marble from it at random. It is black. You then put it back in the bag and then draw another marble from the same bag at random. What is the probability that the second marble is black?
As there have been quite a few probability problems lately, I'll not post another for a while - unless it's excellent.
As there have been quite a few probability problems lately, I'll not post another for a while - unless it's excellent.
Labels: Probability
posted by Chris at
7:58 AM
11 Comments:
Chris, I've spent a couple of minutes to solve it, that's why I don't want to post neither answer, nor solution. But,... have you bought probability theory textbook?
Quantense
Hi quantense. I'd be happy for you to post the result (without an explanation) and appreciate that you would like to give everyone else a chance to work it out for themselves, before giving the reasons for the result.
The only book I've bought lately is "The Road To Reality" by Roger Penrose. I didn't buy it as a source of problems to post though.
I got all these probability questions by trawling the internet. I only post them when I can provide my own explanation and am completely confident in the result (i.e. I have at least two ways of doing it). Greetz.
OK, Chris, it's 5/8.
Roger Penrose?! Great!
I've read "An Emperor New Mind",
his way of providing things is very beautiful. Of course he's written lots of his own ideas there. The one I like very much is about comparing the roles of R-operation and black hole information lost.
Hi quantense. Phew! you've got the same result as me. I read emperor a few years ago. I can't remember it clearly. But I do like his style in his popular books. I've never seen any of his serious academic work. I expect it's beyond my reach in my available time.
this sounds right(i think)
bag A:1 black,3 white
bag B:3 black,1 white
for bag A,chances for drawing the first black marble would be 1 in 4.To repeat this would be another 1 in 4.Probability of having a black marble twice is 1/16.(1/4 x 1/4 = 1/16)
for bag B,you would have a 3 in 4 shot at a black marble.doing it again can read-3/4 x 3/4 = 9/16.
since it's a 50/50 flip on which bag you have,you must split the difference.Take half the probability,of having 2 black marbles in a row,from bag A (0.5/16),and add it to half the probability from bag B (4.5/16).
so-5/16 are the chances of 2 black marbles(or white,for that matter) in a row.But since we know the first one is black,our odds double.(5/16) x 2 = 5/8
if this is wrong,it's cuz i'm a little drunk right now
Hi Knightmare. It seems the alcohol has fixed your broken brain ;) Was that a lucky guess about doubling the odds because the first marble was black? Whatever, you got it right.
I deleted your comment about the post going through (the power I wield is awesome;)). The post not going through seems to happen quite often. Posting another comment seems to help kick stuck ones through. Greetz.
Generally, P(x│y) = P(x & y)/P(y) (│ => "given that")
In this case both x and y correspond to drawing a black marble.
P(B & B) = ½(¾*¾) + ½(¼*¼) = 5/16
P(B) = ½*¾ + ½*¼ = ½ (can see that directly as well)
So, P(B│B) = P(B & B)/P(B) = (5/16)/(½) = 5/8
wow, u uys did it the hard way, all you have to do is multiply the probability of the first bag and the second bag, then time by two, because you put the marble back and did it again. it should come out as 6/32. =D if im wrong, let me no. jesusfreakalloftheway@yahoo.com thanxs
jesus.... The answer is 5/8. I don't understand your explanation at all.
A:3/4
B:1/4
Anonymous. A few words of explanation would have been helpful. I've no idea what your post means.
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