A small-business Web site contains 100 pages and 60%, 30%, and 10% of the pages contain low, moderate, and high graphic content, respectively. A sample of four pages is selected without replacement, and X and Y denote the number of pages with moderate and high graphics output in the sample. Determine: a. fxy(x, y) b. fx(x) c. E(X) d. fyß(y) e. E(Y | X = 3) g. Are X and Y independent?

Step-by-step explanation:

Given that:

A small-business Web site contains 100 pages and 60%, 30%, and 10%  of the pages contain low, moderate, and high graphic content, respectively.

. A sample of four pages is selected without replacement,

Let  X and Y denote the number of pages with moderate and high graphics output in the sample

We are meant to determine

a)    from the given data in the question;

However; the probability mass function can be expressed via the relation:

We can now have a table shown as :

               0           1               2              3              4          Total    

0              0.1244     0.0873     0.02031    0.0018     0.0001   0.234

1               0.2618     0.13542   0.02066    0.00092    0         0.419

2              0.1964     0.0666    0.00499       0              0         0.268

3              0.0621     0.01035     0                 0              0         0.073

4              0.0069       0              0                0              0          0.007

Total   0.6516   0.2996   0.0460      0.0028  0.0001    1

b)

The marginal distribution definition of

From the table above ; the corresponding values of  are :

X           0           1         2          3           4

   0.234   0.419  0.268   0.073    0.007

( since represent the vertical column)

c) E(X)

By using the expression

we have:

E(X) =

E(X) = 0 + 0.419 + 0.536 + 0.218 + 0.028

E(X) = 1.202

d) fyß(y)

Using the thesis of conditional Probability; we have :

The conditional probability for the mass function is then:

where;

values of for every y ∈ (0,1,2,3,4)

Therefore; the mass function is:

e) E(Y | X = 3)

By using the expression

we have:

⇒

= 0.143

The value of E(Y | X = 3) = 0.143

g) Are X and Y independent?

To Check if X and Y independent; Let assume if ; then we can say that X and Y are independent.

From the above previous table :

= 0.1244 + 0.087268+0.02031+ 0.001836 + 0.0001

 = 0.234

We conclude that ; As such X and Y are said to be  non - independent.

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.