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flutherother's avatar

What is the probability of drawing a white counter?

Asked by flutherother (28630points) January 6th, 2011

A bag contains one counter, known to be either white or black. A white counter is put in, the bag shaken, and a counter drawn out, which proves to be white. What is now the chance of drawing a white counter?

Can anyone solve this problem?

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12 Answers

CyanoticWasp's avatar

The chance that the original counter was white is 50%. When a white counter was placed in the bag and then a white counter withdrawn, then you’re back to the original 50–50 chance of the ‘original’ counter being white.

6rant6's avatar

If you assume that the probability of the first counter being either white or black is equal, then on the first draw, you will get white ¾ of the time: 2/4 when they were both white and ¼ wwhen you got the single white one. So your chances of getting a second white one are (2/4 out of ¾) or ⅔.

PhiNotPi's avatar

Here is a complete solution. Since one counter could be either black or white, with the other known to be white, there are two possible contents of the bag: BW and WW. From this you can tell that there is a 50% chance of a 50% chance, and a 50% chance of a 100% chance. There is a 25% chance that the bag has BW and a white is drawn, and a 50% chance that the bag contians a WW and a white is drawn. Add these together and you get a 75% chance (¾ths) chance of drawing a white counter.

flutherother's avatar

@CyanoticWasp That is what you would expect but it isn’t true. @6rant6 has the right answer though I’m not sure I follow the logic

6rant6's avatar

@PhiNotPi If you want to solve it your way, you need to start out with four cases, not two.

Here they are:
WW1 (two whites in the bag, first counter selected)
WW2
BW1
BW2

Now, once we have the first counter, we can eliminate BM1 from the set, since we didn’t draw the first (Black) counter. In the remaining three cases, two of them will yield a second white counter.

submariner's avatar

Extra credit:

Suppose we have a bag with 2 white counters and 1 black. In that case, we say that chance of a randomly drawn counter being white is ⅔.

We can also say, following 6rant6, that the chance of a white counter being drawn in the OP’s case is also ⅔.

Are we working with the same concept of probability in these two cases? Why or why not?

6rant6's avatar

I’m not quite sure I have the problem… Are you saying that IF the first player draws white, then the second player draws…

That answer would be white half the time, not ⅔.

If you are saying that the first player draws a counter, and then the second player draws a counter, then they each have ⅔ chance of having white. Or ⅓ chance of both having white.

submariner's avatar

Sorry, I was unclear. I just meant an simple example of probability, where you know before drawing that 2 counters are white and 1 is black, and draw once, as opposed to the OP’s more complex case.

markferg's avatar

submariner has used quantum theory to explain the problem.

The first counter can be either black or white but no one knows. In quantum theory, we would say that there is a superposition of both black and white counters, they both exist inside the bag at the same time. Now add a white counter so, quantum mechanically, we have 1 black and 2 whites inside the bag. Thus, ⅓ chance of pulling out a black, ⅔ chance of pulling out a white.

submariner's avatar

Thanks markferg. That sheds some light on something else I’ve been working on, which concerns the (mis)application of probability theory to the issue of free will.

I’m still not sure the statement “the probability of drawing a white counter is ⅔” means the same thing in my example and the OP’s example. I’ll have to mull it over a bit more.

PhiNotPi's avatar

I misread the problem. I thought the question wanted to find the probabilty of drawing a white counter, not drawing two whites in a row. Here is my modified answer:

You know that there are three whites in all, one if the bag was BW, and two if the bag was WW. Since we know that one white was drawn, there is a ⅓ chance that it was the white from the BW bag, and there is a ⅔ chance that it is from the WW bag. If it was from the BW bag, there is a 0% of drawing a second white. If it was the WW bag, there is a 100% chance of drawing another white. There is a ⅓ chance of a 0% chance, and a ⅔ chance of a 100% chance. This makes a ⅔ chance of drawing a second white.

PhiNotPi's avatar

@submariner Yes we are using the same laws of probabilty. When we have a probabilty of probabilties situation similar to this, we can just add the bags together. This only works when the chance of each different situation is the same. If the unknown counter had a ⅔ chance of being white, there would be a slightly more complicated process than just adding them together. The BW and WW bags make a BWWW bag. Since we know that one white is removed, the bag becomes BWW. There is now a ⅔ chance of drawing a second white.

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