# Suppose that 3 balls are chosen without replacement from an urn ..without replacement

So this is a two part question, after working the problem I got the first part correct, after completing and checking part 2, I did not get the correct answer, i see the correct method but still not sure how why, my method is not correct

Part 1 of the question

$Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let Xi equal 1 if the ith ball selected is white, and let it equal 0 otherwise. Give the joint probability mass function of (a) X1 , X2$

$$P(0,0) = 8/13*7/12$$ $$P(0,1) = 8/13 *5/12$$ ...etc

I get the first part and how the probability is determined

Second part of the question below

$In part 1, suppose that the white balls are numbered, and let Yi equal 1 if the i th white ball is selected and 0 otherwise. Find the joint probability mass function of (a) Y1 , Y2, Y3$

Correct Solution/Method $$P(0,0,0) = (10*9*8)/(13*12*11)$$ $$P(0,1,0) = 1/13(10*9/12*11) *3$$

assuming is the white ball labelled 1

My method

$$P(0,0,0) = 8/13*7/12*6/11$$

as here I am assuming the probability that the ball chosen is not white, like in part one of this question

$$P(0,1,0) = (1/13)8/13*7/12*3$$

assuming the white ball labelled 1 is chosen

So i get the 1 in 13 chances of choosing the white ball labelled (1) and multiply by 3 for the number of ways ball can be chosen, what throw me off is the 11*10*9/13*12*11 why not

The main difference in the second problem is that you can still pick out a white ball on the $i$-th turn and still have $Y_i=0$. For example, if on the first turn you remove the white ball numbered $2$ then you will still have $Y_1=0$ (same if you get balls numbered $3,4,5$). This means there are more balls for you to choose from. Working out $P(0,0)$ we see that in our first pick we can pick any ball as long as it is not numbered $1$ or $2$ (since $Y_1=0, Y_2=0$) so we have $11/13$ probability at first ($8$ black balls plus white balls numbered $3,4,5$) and then $10/12$ and eventually $9/12$ as we continue to remove balls without the numbers $1,2$. Hopefully that makes sense and you can figure out why the $P(1,0,0)$ case is also different.
• In the setup for the second problem the time at which a ball is picked is irrelevant. So, for example, $Y_1=1$ as long as the first ball is picked, regardless of whether it was the first or second one to be picked etc. – Twis7ed Apr 13 '17 at 13:27