There are $2$ balls in an urn. Each ball is either white or black. If a white ball is put into an urn and thereafter a ball is drawn at random from the urn, then find the probability that it is white.

Does "each ball is either white or black" mean that there are 3 cases, which are

  • $2$ black balls

  • $2$ white balls

  • $1$ black ball and $1$ white ball?

Also if we add one white ball and then find probability for three cases and then add three probabilities , wouldn't probability be greater than 1?


As you say, there are three cases, as the original urn has $U$ white balls where $U$ could be $0$, $1$ or $2$. We can calculate the probability the event of drawing a white ball, $W$ in each case. We would write these as $\mathbb P(W\mid U=0)$, $\mathbb P(W\mid U=1)$ and $\mathbb P(W\mid U=2)$. As Edward Porcella says, we have $\mathbb P(W\mid U=0)=1/3$, $\mathbb P(W\mid U=1)=2/3$ and $\mathbb P(W\mid U=2)=1$.

We don't simply add these probabilities. First we have to multiply each probability by the probability of being in that case in the first place. This is because $$\mathbb P(W\mid U=0)\mathbb P(U=0)=\mathbb P(U=0\text{ and }W),$$ and so on, and $$\mathbb P(W)=\mathbb P(U=0\text{ and }W)+\mathbb P(U=1\text{ and }W)+\mathbb P(U=2\text{ and }W).$$

Unfortunately the original problem doesn't tell us what $\mathbb P(U=0)$, etc, are.

The most natural interpretation is that each ball is independently equally likely to be black or white. This gives $\mathbb P(U=0)=\mathbb P(U=2)=1/4$ and $\mathbb P(U=1)=1/2$, and so the answer is $2/3$. Edward Porcella's answer makes the (different) assumption that all three cases are equally likely. This gives the same answer, as in fact does any interpretation where the cases $U=0$ and $U=2$ are equally likely.

  • $\begingroup$ I think I agree, it's more natural to assume each ball is independently likely to be black or white, which yields a different calculation, although the result is the same. $\endgroup$ – Edward Porcella Apr 8 '17 at 22:12

The probability is $\frac23$. As you say, after the white ball is added there are one, two, or three white balls in the urn. But the probability of drawing a white ball is not greater than $1$ since only one of the three possibilities is true, and we don't know which. You're either certain to draw a white ball, or have a two-thirds chance, or a one-third chance. The average of $\frac13$,$\frac23$, and $\frac33$ is $\frac23$.


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