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 $

The answer is given by

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

  • $\begingroup$ I am a newbie here. Don't understand why Y1=0 doesn't mean 12/13. And suppose in first pick ball no. 2 is picked then for the second pick, whatever we pick will result in score of 0 as ball no. 2 is not there anymore and that's the only ball that can give a score of 1 on second pick. $\endgroup$ – Prakash - Crow Canyon Apr 13 '17 at 10:53
  • $\begingroup$ 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. $\endgroup$ – Twis7ed Apr 13 '17 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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