# Put a white ball into a basket, then take out a white ball. Is the remaining ball white?

It seems to be very simple question, but I'm afriad that I am missing something.

Ok, so you have a basket with 1 ball- which can be black or white with an equal proability. Then you insert to the basket white ball, and by coincidental selection you take out white ball out of the basket.

What is the probability that the remaining ball is white?

Thanks

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Excellent question. A quintessence of a problem on Bayes posterior probability. Total probability for evidence to happen (for white ball to be removed): $$P(E)=\frac{1}{2}\cdot 1+\frac{1}{2}\cdot \frac{1}{2}=\frac{3}{4}$$ Prior hypothesis probability (for the first ball to be white): $$P(H)=\frac{1}{2}$$ The probability for the evidence to happen if the hypothesis is true (for white ball to be removed if the first one was white): $$P(E|H)=1$$ Posterior probability for the hypothesis $H$ in the presence of the evidence $E$: $$P(H|E)=\frac{P(E|H)\cdot P(H)}{P(E)}=\frac{1 \cdot \frac{1}{2}}{\frac{3}{4}}=2/3$$

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There are four equally likely possibilities:
a) Ball 1 is white, and you remove ball 1 from the basket.
b) Ball 1 is black, and you remove ball 1 from the basket.
c) Ball 1 is white, and you remove ball 2 from the basket.
d) Ball 1 is black, and you remove ball 2 from the basket.

We rule out case b), since the removed ball was white. The other three cases are equally likely, and in two of them (a) and c)) the remaining ball is white. The answer is thus 2/3.

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One can also use that fact that the probability the remaining ball is white is $$P(\text{remaining ball is white}) = 1-P(\text{remaining ball is black})$$

$$= 1- \frac{1}{3} = \frac{2}{3}$$

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