0
$\begingroup$

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

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Alex G., Watson, John B, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Q4) at second stage $2$ balls are extracted so that $Y<3$ a.s. making $P(X=x\mid Y=3)$ meaningless. $\endgroup$ – drhab Jun 12 '16 at 15:49
  • $\begingroup$ I should pay more attention on reading (y) $\endgroup$ – Paul Jun 13 '16 at 13:09
1
$\begingroup$

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:

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.