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
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$2$ black balls
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$2$ white balls
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$1$ 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?
Best Answer
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.