Originally posted by AThousandYoungI figured you'd see that after working it out completely yourself.
Wait, this seems to give me a 67% chance. Hmmm.
I think my mistake was not realizing black/black has two out of two possible ways to pick black first, while white/black has one out of two possible ways to pick black first. If I did not know the ball I picked was black my 50% would be correct I think.
What part of my explaination doesn't clearly show what's happening?
1 edit
Originally posted by wolfgang59okay -- we have two balls -- the first ball has a 50% chance of being wb (white) or bw (black), the 2nd ball is just plain old B (black)
A bag contains a ball which is known to be either black or white (even chance). A black ball is added. The balls are then drawn from the bag one at a time.
What is the probability of the second ball being black
1. if the first is black?
2. if the first is white?
3. if the colour of the first is unknown?
there are 4 possibilities when drawing the two balls from the bag
1.wb-B
2.bw-B
3.B-bw
4.B-wb
if the first ball is black, we eliminate scenario #1 - of the remaining three, the second ball is black in 2 of them -- so the probability is 2/3
if the first ball is white, the only possible scenario is #1 - so the second ball must be black -- so the probability is 1/1
if the first ball is unknown, then all four scenarios are possible -- the second ball is black in 3 of them -- so the probability is 3/4