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?
My question is: Can I find the probability that the fourth ball is black if the other three balls are all white and then count $ 1- $ this probability?
I solved my question. For the exercise, that's my trying:
$E$ = the fourth ball is black
$A$ = the three balls are ALL WHITE
Then $P(E) - P(E|A)$ will give me what I need.
$C$ = 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 $P(E|D)$.