# Probability of drawing of a 3rd white ball after two other white balls have already been drawn from a bag containing only black and/or white balls

A bag holds four balls. One of them is white. Each of the others is either black or white at equal chance. You randomly draw out two balls and discover they are both white. If you then randomly draw a third ball, what is the chance that it is white?

I've seen answers varying from 0.27 through 0.58 to 0.75. What is the correct answer?

• Welcome to stackexchange. Can you show us your attempt? With just four balls you should be able to write the whole tree of possibilities. Feb 2, 2018 at 16:05
• I feel this is a tough question, where interpretations can really cause a mix of answers, just my thoughts. Feb 2, 2018 at 20:13

The probabilities that we have $(1,2,3,4)$ white balls in the bag are $$\left({1\over8},{3\over8},{3\over8},{1\over8}\right)\ .$$ Given that we have $(1,2,3,4)$ white balls in the bag the probabilities that we draw two white balls are $$\left(0,{1\over6}, {1\over2},1\right)\ .$$ The overall probability $p(2w)$ to draw two white balls then is given by $$p(2w)={1\over8}\cdot0+{3\over8}\cdot{1\over6}+{3\over8}\cdot{1\over2}+{1\over8}\cdot1={3\over8}\ .$$ Finally the probability that all three balls are white computes to $$p(3w)={1\over8}\cdot0+{3\over8}\cdot0+{3\over8}\cdot{1\over4}+{1\over8}\cdot1={7\over32}\ .$$ It follows that the conditional probability $p(3w|2w)$ is given by $$p(3w|2w)={p(3w\wedge 2w)\over p(2w)}={p(3w)\over p(2w)}={7\over12}=0.58333\ldots\ .$$
• @imranfat: $p(3w)$ is the probability that all three drawn balls are white, and this is equal to the probability that the first two balls are white as well as all three. Check how I computed $p(3w)$. Feb 2, 2018 at 20:16