# Probability boxes with balls [duplicate]

I have two boxes, one with ten balls, eight white and two black and the other with ten balls, four white and six black. Without seeing I choose a box and choose three balls from this. What is the probability that the fourth ball I'll choose is black if the other three are not all white?

$$E$$ = the fourth ball is black

$$A$$ = the three balls are ALL WHITE

Then $$P(E) - P(E|A)$$ will give me what I need.

I think I am ok with $$P(E|A)$$.

For $$P(E)$$:

$$C$$ = I choose first box and $$D$$ = I choose second box

$$P(E) = P(E|C) \cdot P(C) + P(E|D) \cdot P(D)$$

I will find $$P(E|C), P(E|D)$$ from the right groups of four balls

So, if $$a$$ is a white ball and $$b$$ is a black ball then:

For the first box: $$aaab, baab, abab, aabb$$ I sum the possibilities for all these groups and I get $$P(E|C)$$.

For the second box: $$aaab, baab, abab. aabb, bbab, babb, abbb, bbbb$$ I sum the possibilities for all these groups and I get $$P(E|D)$$.

The other thing I thought was: $$E$$ = the fourth ball is black

$$B$$ = the three balls are NOT all white

Then $$P(E|B)= \frac{P(E \cap B)}{P(B)}$$

• Note: I'm not sure why the prior version of this question was closed...it seemed to be drawing a good discussion. Perhaps it was edited post-closing? In any case, I have voted to re-open it. – lulu Apr 25 '20 at 15:50
• @lulu: Concur; it is now open again. – Brian M. Scott Apr 25 '20 at 15:59