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

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  • $\begingroup$ 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. $\endgroup$ – lulu Apr 25 '20 at 15:50
  • $\begingroup$ @lulu: Concur; it is now open again. $\endgroup$ – Brian M. Scott Apr 25 '20 at 15:59