Question

Suppose three balls are chosen from a ballot box containing 3 red balls, 4 white balls and 5 blue balls. Let X be the number of red balls and Y the number of white balls. Make a joint distribution table of X and Y and indicate the marginal probabilities.

*STEP BY STEP*

Answer 1

A ballot box contains 3 red balls, 4 white balls and 5 blue balls.
Three balls are chosen from the ballot box.
X denotes the number of red balls and Y denotes the number of white balls. X and Y can take values 0,1,2,3.
[3 blue balls, 0 red and 0 white balls are chosen]






[Since, 3 balls are chosen, 1 red and 3 white balls can never be chosen.]






Therefore, the joint distribution is given by

X Y	0	1	2	3	P(Y=y)
0	1/22	3/22	3/44	1/220	14/55
1	2/11	3/11	3/55	0	28/55
2	3/22	9/110	0	0	12/55
3	1/55	0	0	0	1/55
P(X=x)	21/55	27/55	27/220	1/220	1

Each cell value (except last row and last column) represents the joint probability of X=x_i and Y=y_j
The last row  represents the marginal probability distribution of X=x_i and the last column represents the marginal probability of Y=y_j.


I hope this answer will help you to a great extent. Thank You. :)