Question

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation.
Y
p(x,y), 0 1 2
0 .10 .04 .02
x 1 .08 .20 .06
2 .06 .14 .30
a. What is P(X = 1 and = 1)?
b. Compute P(X land Y 1).
c. Give a word description of the event {X t- 0 and Y 0}, and compute the probability of this event
d. Compute the marginal pmf of X and of Y. Using pX(x), what is P(X 5 1)?
e. Are X and Y independent rv's? Explain.

Answer 1

Answer:

Step-by-step explanation:

                              Y

p(x,y)            0          1            2

           0      0.10     0.04     0.02

x          1       0.08     0.2     0.06

            2       0.0      0.14     0.30

a) What is P(X = 1 and = 1)

From the table above we have

P(1,1) = 0.2

b)  Compute P(X ≤ 1 and Y ≤ 1).

$=p(0,0)+p(0,1)+p(1,0)+p(1,1)\\\\=0.1+0.04+0.08+0.2\\\\=0.42$

C)

Let A ={X ≠ 0 and Y ≠ 0}

p{X ≠ 0 , Y ≠ 0}

= p(1,1) + p(1,2) + p(2,1) + p(2,2)

= 0.20 + 0.06 + 0.14 + 0.30

=0.7

d) The possible X values are in the figure 0,1,2

$p_x(0)=p(0,0)+p(0,1)+p(0,2)\\\\=0.1+0.04+0.02\\\\=0.16\\\\p_x(1)=p(1,0)+p(1,1)+p(1,2)\\\\=0.08+0.2+0.06\\\\=0.34\\\\p_x(2)=p(2,0)+p(2,1)+p(2,2)\\\\=0.06+0.14+0.3\\\\=0.5$

The possible Y values are in the figure 0,1,2

$p_y(0)=p(0,0)+p(1,0)+p(2,0)\\\\=0.1+0.08+0.06\\\\=0.24\\\\p_y(1)=p(0,1)+p(1,1)+p(2,1)\\\\=0.04+0.2+0.14\\\\=0.38\\\\p_y(2)=p(0,2)+p(1,2)+p(2,2)\\\\=0.02+0.06+0.3\\\\=0.38$

So the probability of x ≤ 1 is

$p(x\leq 1)=p_x(0)+p_x(1)\\\\=0.34+0.16\\\\=0.50$

e) From the table

$p_x(x=1,y=1)=p(1,1)\\\\=0.2\\\\p_x(1)=0.34\\\\p_y(1)=0.38$

we multiply both together

0.34  x 0.38

=0.1292

Therefore p(1,1) is not equal px(1), py(1)

Hence x and y are not independent it is not equal