Need help solving this problem

[José Silva]

A box contains 4 parts, two of them are defective . To identify the defective parts, successive extractions of a part are carried out, without repositioning. Let X and Y be the real random variables that represent, respectively, the number of extractions performed to detect the first defective part and the number of additional extractions required to detect the second defective part.

a)Check that:

P(X=x, Y=y)=1/6, (x,y) ∈ S={(x,y)∈ {1,2,3}×{1,2,3}:x+y≤4}.

b)Calculate the probability of being needed

i)fewer extractions to detect the second defective part than to detect the first;

ii)at least 3 extractions to detect the two defective parts.

c)Show that X and Y follow the same law of probability.


d)Determine Cºv(X,Y).

e)Construct the probability function of X conditioned by Y = 2 and calculate E (X / Y = 2).


Sorry if its poorly translated, the original problem wast in english and i had to translate it.

 

[Steven G]

Welcome to FMH!

On this forum we prefer that the student solves their own problem with the assistance of forum helpers.
What this means is that you need to be actively involved in solving your problem. So please post the work which you have done so far so we can see how you want to solve it and where you might be getting stuck. Then we can help you progress to the final answer.

 

[José Silva]

I have solved all the lines except for c(its the one i really need help), i can post my work here, but its on portuguese.

 

[Steven G]

Can you tell us how You define Cºv(X,Y)??

 

[José Silva]

Cºv(X,Y)=E(X,Y)-E(X)E(Y)=(5/2)-(25/9)=-5/18