Question

Two individuals were asked to pick a number from 1 to 3 S = {1, 2, 3] (with replacement). Let X be the sum of the two numbers chosen and let Y be the product of the two numbers chosen (5 marks)
a) Construct the joint probability distribution table for (X, Y).
b) Calculate ??(??=4|??=4).
c) Are X and Y independent? Explain!

Answer 1

a)

The sample space of picking two numbers by two persons {(1,1) , (1,2) , (1,3) , (2,1) , (2,2) , (2,3) ,(3,1) , (3,2) , (3,3) }

Based on this sample space, the possible values of X and Y are

X = (2, 3,4,3,4,5,4,5,6)

Y = (1,2,3,2,4,6,3,6,9)

So, sample space of (X, Y) are,

(2,1) , (3,2) (4,3) (3,2) (4,4) (5,6) (4,3) (5,6) (6,9)

Size of sample space = 9

Elements (3,2)(4,3) (5,6) occurs twice, all other elements occur once

The joint probability distribution table for (X, Y)

X \ Y	1	2	3	4	6	9
2	1/9	0	0	0	0	0
3	0	2/9	0	0	0	0
4	0	0	2/9	1/9	0	0
5	0	0	0	0	2/9	0
6	0	0	0	0	0	1/9

b)

From the table,

P(Y = 4) =  P(X = 2, Y = 4) +  P(X = 3, Y = 4) +  P(X = 4, Y = 4) +  P(X = 5, Y = 4) +  P(X = 6, Y = 4) = 1/9

P(X = 4 | Y = 4) = P(X = 4, Y = 4) / P(Y = 4)  = (1/9) / (1/9) = 1

c)

P(X = 4) =  P(X = 4, Y = 1) + P(X = 4, Y = 2) + P(X = 4, Y = 3) + P(X = 4, Y = 4) + P(X = 4, Y = 6) + P(X = 4, Y = 9)

= (2/9) + (1/9) = 1/3

P(X = 4) * P(Y =4) = (1/3) * (1/9) = 1/27

So, P(X = 4) * P(Y =4) $\ne$ P(X = 4, Y = 4)

Hence  P(X = x) * P(Y =y) $\ne$ P(X = x, Y = y) does not hold for all values of X and Y.

and X and Y are not independent.