Two random variables X and Y are independent. Each has a binomial distribution with success probability 0.4 and 2 trials.

(a) Find the joint probability distribution function f(x,y).
(b) Give the joint probabilities using a table. Hint, the size of the tables is 3 by 3.

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Answer:

Step-by-step explanation:

Given that two random variables X and Y are independent. Each has a binomial distribution with success probability 0.4 and 2 trials.

When x and y are independent joint probability would be product of individual probabilities

pdf of X

X is Binom (2,0.4)

and Y is Binomi (2,0.4)

Hence joint distribution of XY would be

P(X=x, Y=y) =[tex]2Cx (0.4)^x (0.6)^{2-x} *2Cy (0.4)^y (0.6)^{2-y}[/tex]

for x=0,1,2 and y =0,1,2

b) Joint probability using table

PDF of X is

X        0            1           2

p       0.36    0.48      0.16

and same for Y also

Joint prob would be

X  Y       0            1              2

0      0.1296     0.1728      0.0576

1       0.1728      0.2304     0.0768

2      0.0576     0.0768     0.0256

Joint probability distribution function are used to represent the probability of multiply variables

The joint probability distribution function is [tex]f(x,y) = ^2C_x *0.4^x * 0.6^{2- x} *^2C_y * 0.4^y * 0.6^{2- y}[/tex]

The given parameters are:

[tex]p = 0.4[/tex] --- the probability of success

[tex]n = 2[/tex] ----the number of trials

The joint probability distribution function f(x,y) is calculated as:

[tex]f(x,y) = ^nC_x * p^x * (1 -p)^{n- x} *^nC_y * p^y * (1 -p)^{n- y}\\[/tex]

So, we have:

[tex]f(x,y) = ^2C_x *0.4^x * (1 -0.4)^{2- x} *^2C_y * 0.4^y * (1 -0.4)^{2- y}[/tex]

Evaluate the differences

[tex]f(x,y) = ^2C_x *0.4^x * 0.6^{2- x} *^2C_y * 0.4^y * 0.6^{2- y}[/tex]

The above represents the joint probability distribution function f(x,y)

When x = 0, y = 0;

We have:

[tex]f(0,0) = 0.130[/tex]

When x = 0, y = 1;

We have:

[tex]f(0,1) = 0.173[/tex]

When x = 0, y = 2;

We have:

[tex]f(0,2) = 0.058[/tex]

When x = 1, y = 0;

We have:

[tex]f(1,0) = 0.173[/tex]

When x = 1, y = 1;

We have:

[tex]f(1,1) = 0.230[/tex]

When x = 1, y = 2;

We have:

[tex]f(1,2) = 0.077[/tex]

When x = 2, y = 0;

We have:

[tex]f(2,0) = 0.058[/tex]

When x = 2, y = 1;

We have:

[tex]f(2,1) = 0.077[/tex]

When x = 2, y = 2;

We have:

[tex]f(2,2) = 0.026[/tex]

So, the joint probability as a table is:

X /Y       0            1              2

0      0.1296     0.1728      0.0576

1       0.1728      0.2304     0.0768

2      0.0576     0.0768     0.0256

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