A bag contains one counter, whose color is unknown but....

[21385]

A bag contains one counter, whose color is unknown but is either black or white. 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 the white counter from the bag?

I am pretty sure that it is not 50/50.

Thanks

 

[Denis]

21385 = BACHE ?

Hint: chances of pulling original counter 1st = 50%; chances that it is white = 25%

 

[21385]

I am still not sure...can someone explain it to me? thanks

I am thinking that since you're putting a white counter in and taking that a white counter out, you can either taking the original or the new one out. (50/50) If you took the original out, in which the possibility of being white is 50%, so in total 25%, you are guaranteed a white counter in the end. If you are taking the new one out, which is 50%, you are left with a 50% chance of getting a white counter for your original counter, thus the combined percentage is 25%. So you add these together and get 50%. I am pretty sure my thinking is wrong but can someone explain it to me? Thanks

 

[pka]

The probability is 2/3.
Now I have simply given you the answer without explaining it.
But you posted this under a basic ‘Algebra Forum’, which means I don’t know how much you know about Bayes Theorem.

This is the same problem as the Monty Hall problem. As the three card problem: one has two red faces, one has two blue faces and one has a face of each. You can do a web search of more on this.