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?
(In reply to
Without replacement (spoiler) by Steve Herman)
What if I turn your argument around and ask what is the probability that the new counter was drawn out? Taking a look, your argument has 2 out of 3 ways to draw the new counter so by the logic that that each scenario is equally likely then the probability is 2/3. This is clearly false as there are exactly two counters in the bag and they are both equally likely to be drawn.
So something is wrong, specifically that each of the three scenarios have an equal chance of occurring. There is an equal chance of drawing the original counter vs the new counter. This then implies that scenario c) has a 1/2 chance of happening. That leaves each of scenario a) and b) a 1/4 chance each. Thus the probability of drawing a white counter is 3/4.
This problem is equivalent to being given a bag with two coins, one is normal heads/tails and the other is double heads. And then being asked what is the probability that heads shows up when one coin is chosen at random and flipped. There are three heads and one tail, so the probability is 3/4.
re: Without replacement (spoiler)
