Probability in a Urn Modell If we have a Urn, with 2 black balls,3 red ones and 4 yellow ones.
I'd like to determine the probability of drawing a black ball before drawing a red ball.
So the probabilies to consider are the 3-tupels
(black,red,random),(random,black,red) and (black,yellow,red)
I tried to split the Set of the desired Event into 3-smaller events
namly
${ A }_{ 1 }=\left\{ ({ w }_{ 1, }{ w }_{ 2 },{ w }_{ 3 }):{ w }_{ 1 }\in (1,2),{ w }_{ 2 }\in (3,4,5),{ w }_{ 3 }\in (1,..,9) \right\} \\ { A }_{ 2 }=\left\{ ({ w }_{ 1, }{ w }_{ 2 },{ w }_{ 3 }):{ w }_{ 1 }\in (1,...,9),{ w }_{ 2 }\in (1,2),{ w }_{ 3 }\in (3,4,5) \right\} \\ { A }_{ 3 }=\left\{ ({ w }_{ 1, }{ w }_{ 2 },{ w }_{ 3 }):{ w }_{ 1 }\in (1,2),{ w }_{ 2 }\in (6,7,8,9),{ w }_{ 3 }\in (3,4,5) \right\} $
but im struggeling to get the orders of those sets and im also not sure if the number of total outcomes=84.
appreciate any help

So in our draw of $3$ balls, there's definitely a black ball and a red ball. So the third ball can be a yellow or black or red ball. If it's a yellow ball, the favourable cases are just the number of permutations where black comes before red. This is $2$ (YBR,BRY), so the probability is just $1/12$. If it's a black ball, the number of favourable cases is again $2$ (BBR, BRB) and this time the probability is ${2}\choose{2}$$. 3.2$/ ${9}\choose{3}$ the third ball is red, theres just one favourable case,(BRR), and it's probability is $2.$${3}\choose{2}$/${9}\choose{3}$. Since these cases are disjoint, the total probability is just the sum of these probabilities.
