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
 A: 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.
A: The number of ways to draw black-yellow-red (or the number of outcomes in $A_3$) is just $2\times 4\times 3 = 24$.
For black-red-random, you can draw the first ball in $2$ ways, the second one in $3$ ways and the last one in $7$ ways. Total $42$ ways.
For random-black-red, you'll further have to split this into the following $3$ cases: yellow-black-red, black-black-red, red-black-red. For each case, find the number of ways to draw balls of these colors in succession and add them up.
A: Outcomes with exactly one black and one red: BRY, BYR, YBR. Outcomes with 2red, 1 black: BRR, outcomes with1 red 2 black: BBR. The probabilies are $3\times \frac{2}{9}\frac{3}{8}\frac{3}{7} + \frac{2}{9} \frac{3}{8} \frac{2}{7} + \frac{2}{9}\frac{1}{8}\frac{3}{7}$
