# 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

• Does the question specify that you're drawing exactly $3$ balls in succession? – Shirish Kulhari Apr 18 '18 at 8:36
• Yes drawing in succesion and not retuniing the balls to the urn. shout have notd thissorry – johnka Apr 18 '18 at 8:38
• Are you sure that you stop after drawing $3$ balls allready? Are you not going on with drawing until it is clear whether the event (a black ball is drawn before any red balls are drawn) occurred or did not occur? – drhab Apr 18 '18 at 8:55

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.
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.
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 $3\times \frac{2}{9}\frac{3}{8}\frac{3}{7} + \frac{2}{9} \frac{3}{8} \frac{2}{7} + \frac{2}{9}\frac{1}{8}\frac{3}{7}$