Marginal/Conditional Probability Mass Functions

[vikkisut88]

Homework Statement

A bag contains four dice labelled 1,...,4. The die labelled j has j white faces and (6-j) black faces, j = 1,...,4. A die is chosen at random from the bag and rolled. We define X = the number labelling the chosen die.
Y = {0 if the face showing on the die is black; 1 if the face showing on the die is white.

Construct a table displaying the values of the marginal pmf (probability mass function) for X and a separate table displaying the values of the conditional pmf for Y fiven X=x for general x.

Homework Equations

Marginal pmf for X is P_x(x) = P(X=x) = \sum P(x,y)
Conditional pmf for Y given X=x is P_Y/X(y,x) = P(x,y)/P_x(x)

The Attempt at a Solution

Okay so i know the two equations above for the two pmf's. I'm presuming that for the first part (marginal pmf) that you just use x=1,2,3,4 and that P_x(x) is 1/4 for each value of x?
For the second table I'm not so sure as I don't know which y values to put into my table, and how to work out any probabilities as it has to be for a general x and not a given value of x. As at the beginning of the question it states that y=0 or y=1 is it just these two values that i put into my table n then depending on when x=1,...,4 will alter the probability

 

[haruspex]

as it has to be for a general x and not a given value of x

I think they're asking for a 2D table, so for each x you have a distribution for y.

 

[JBGARNER]

X=1, then Pr(Y=0|X = 1)=5/24, Pr(Y=1|X=1)=1/24;
X=2, then Pr(Y=0|X=2)=4/24, Pr(Y=1|X=2)=2/24;
Continue for each value of X.