There are two balls in an urn. Their colors are unknown – they may be black or white. On the first turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.
On the second turn, a ball is drawn from the urn. It is white. It is then replaced into the urn.
We now know that at least one of the balls in the urn is white. We then draw a third time. What is the probability that this ball is white.
To solve this problem – my first instinct is to say that it must be pretty close to fifty percent. It is definitely possible to draw the same ball twice in a row – 25% chance of this. And, there is a 100% chance one ball is white, with a 50% chance the other ball is white as well. How can I set up and solve a problem using this information?
Best Answer
It's necessary to make some assumption about the a priori probability regarding the balls in the urn. In the solution below, I assume that before beginning the draws, each of the two balls could be black or white with equal probability.
Consider $16$ equally likely trials. In $4$ trials both of the urn's balls are black. In $4$ trials both of the urn's balls are white. In $8$ trials the urn has a black ball and a white ball.
In the $8$ trials where the urn has a black ball and a white ball, you will have $2$ trials where you have drawn $2$ white balls. In the $4$ trials where the urn has two white balls, you will have $4$ trials where you have drawn $2$ white balls.
Therefore, there is a $\frac 13$ probability that you are in a trial where the urn has a white ball and a black ball, which means there's a $\frac 16$ probability that the next ball you draw will be black, so the probability that your third draw will be a white ball is $\frac 56$.