# Two stage urn balls without replacement [closed]

I made 3 extractions without replacement from urn with 10 white balls and 2 black balls (first stage), and then another 2 extractions without replacement (second stage). $X$ is a random variable of white balls extracted on first stage, $Y$ is a random variable of white balls extracted on second stage. I need to find:

1. support and probability function for marginal $X$
2. support and probability function for conditional $Y|X=x$
3. support and joint probability of the random vector $(X,Y)$
4. support and probability function for conditional $X|Y=3$

Q1

$P(X=i)=\frac{\binom{10}{i} \binom{2}{3-i}}{\binom{12}{3}}$, $i=1,2,3$

Q2

$P(Y=j|X=0)$ is impossible because on first stage at least 1 white ball is extracted.

$P(Y=j|X=1)=\frac{\binom{9}{j}}{\binom{9}{2}}$, $j=2$

$P(Y=j|X=2)=\frac{\binom{8}{j} \binom{1}{2-j}}{\binom{9}{2}}$, $j=1,2$

$P(Y=j|X=3)=\frac{\binom{7}{j} \binom{2}{2-j}}{\binom{9}{2}}$, $j=0,1,2$

Q3

$P(X,Y)=?$

Q4

$P(X|Y=3)=?$

## closed as off-topic by Alex G., Watson, choco_addicted, John B, LeucippusJun 13 '16 at 0:04

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• Q4) at second stage $2$ balls are extracted so that $Y<3$ a.s. making $P(X=x\mid Y=3)$ meaningless. – drhab Jun 12 '16 at 15:49
• I should pay more attention on reading (y) – Paul Jun 13 '16 at 13:09

Your question can interpreted as: $3$ balls are placed in a binX, $2$ balls in binY and $7$ in binZ.

Let $X$ denote the number of white balls that end up in binX.

Let $Y$ denote the number of white balls that end up in binY.

Let $Z$ denote the number of white balls that end up in binZ.

If $i+j+k=10$ then:

$$\mathbb{P}\left(X=i,Y=j,Z=k\right)=\frac{\binom{3}{i}\binom{2}{j}\binom{7}{k}}{\binom{12}{10}}$$

Leaving $Z$ out you can also write:

$$\mathbb{P}\left(X=i,Y=j\right)=\frac{\binom{3}{i}\binom{2}{j}\binom{7}{10-i-j}}{\binom{12}{10}}$$