Answer:
The joint probility density will be c = 1/36
The joint probability distribution will be c = 1/15
Step-by-step explanation:
For step by step explanation see the pucture attached
Following are the solution to the given question:
To be valid probability functions, the sum of changes across the entire range of x and y variables must equal one.
For point a)
In this case, the cumulative probability density might be represented as follows:
[tex]x=1 \ \ \ \ \ \ \ \ \ \ \ \ \ x=2 \ \ \ \ \ \ \ \ \ \ \ \ \ x=3\\\\[/tex]
[tex]y=1 \ \ \ \ \ \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ \ \ \ \ \ 2c \ \ \ \ \ \ \ \ \ \ \ \ \ 3c\\\\y=2 \ \ \ \ \ \ \ \ \ \ \ \ \ 2c \ \ \ \ \ \ \ \ \ \ \ \ \ 4c \ \ \ \ \ \ \ \ \ \ \ \ \ 6c\\\\y=3 \ \ \ \ \ \ \ \ \ \ \ \ \ 3c \ \ \ \ \ \ \ \ \ \ \ \ \ 6c \ \ \ \ \ \ \ \ \ \ \ \ \ 9c\\\\[/tex]
As just a result, the overall number of probability equals
[tex]= c (1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 + 9) = 36c[/tex]
The sum must now equal one of these for it to be legal. As a result,
[tex]36c = 1 \\\\c = \frac{1}{36}[/tex]
For point b)
In this case, the joint probability could be represented as follows:
[tex]x= -2 \ \ \ \ \ \ \ \ \ \ \ x=0 \ \ \ \ \ \ \ \ \ \ \ x=2 \\\\[/tex]
[tex]y=-2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4c \\\\y=3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3c \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ c[/tex]
As a consequence, the total likelihood in this case is:
[tex]\to c ( 2 + 4 + 5 + 3 + 1) = 15 c\\\\\to 15c = 1 \\\\c = \frac{1}{15}[/tex]
Learn more:
brainly.com/question/10268144
