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.

a. What is P(X = 1 and Y = 1)?

b. Compute P(X â‰¤ 1 and Y â‰¤ 1).

c. Give a word description of the event {X â‰ 0 and Y â‰ 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 â‰¤ 1)?

e. Are X and Y independent rv's? Explain.

[tex]p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+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+0.30=0.50[/tex]

Marginal probability distribution of y:

[tex]p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+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[/tex]

e) The two variables are dependant.

Step-by-step explanation:

The joint pmf of X and Y appears in the accompanying tabulation (attached).

a) Is the proability that one hose being used in the self service island and one hose being use in the full service island.

[tex]P(X=1 \& Y=1)=P(1,1)=0.2[/tex]

(according to the tabulation)

b)

[tex]P(X\leq1 \& Y\leq1)=P(0,0)+P(1,0)+P(0,1)+P(1,1)\\\\P(X\leq1 \& Y\leq1)=0.10+0.08+0.04+0.20\\\\P(X\leq1 \& Y\leq1)=0.42[/tex]

c) This event means that at least one hose is in use in each of the island.

[tex]P(X\neq0 \& Y\neq0)=P(1,1)+P(1,2)+P(2,1)+P(2,2)\\\\P(X\neq0 \& Y\neq0)=0.20+0.06+0.14+0.30=0.70[/tex]

d)

Marginal probability distribution of x:

[tex]p_x(0)=p(0,0)+p(0,1)+p(0,2)=0.10+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+0.30=0.50[/tex]

Marginal probability distribution of y:

[tex]p_y(0)=p(0,0)+p(1,0)+p(2,0)=0.10+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[/tex]

Compute P(X≤1)

[tex]P(X\leq1)=p_x(0)+p_x(1)=0.16+0.34=0.50[/tex]

e) For X and Y to be independent variables, the following conditions must be true:

P(x|y) = P(x), for all values of X and Y.

P(x ∩ y) = P(x) * P(y), for all values of X and Y.

We evaluate the second probability for X=1 and Y=1.

[tex]P(x=1 \cap y=1) = 0.2\\\\p_x(1)*p_y(1)=0.34*0.38=0.1292\\\\\\P(x=1 \cap y=1)\neq p_x(1)*p_y(1)[/tex]

This condition is not satisfied, so the two variables are dependant.