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. yp(x, y) 0 1 2x 0 0.10 0.03 0.01 1 0.08 0.20 0.06 2 0.05 0.14 0.33 (a) Given that X = 1, determine the conditional pmf of Y�i. e., pY|X(0|1), pY|X(1|1), pY|X(2|1). (Round your answers to four decimal places.) y 0 1 2pY|X(y|1) (b) Given that two hoses are in use at the self-service island, what is the conditional pmf of the number of hoses in use on the full-service island? (Round your answers to four decimal places.) y 0 1 2pY|X(y|2) (c) Use the result of part (b) to calculate the conditional probabilityP(Y ? 1 | X = 2). (Round your answer to four decimal places.) P(Y ? 1 | X = 2) = (d) Given that two hoses are in use at the full-service island, what is the conditional pmf of the number in use at the self-service island? (Round your answers to four decimal places.) x 0 1 2pX|Y(x|2)

a) 0.2353; 0.5882; 0.1765

b) 1.0000

c) 0.3654

d) 1.0000

Step-by-step explanation:

This question is not clear. The table is disorganised. The complete question has been added as an attachment.

There is a little difference in the values of the attached file:

At x =2 , y= 0, the value is 0.05 not 0.06

At x =2 , y= 2, the value is 0. 33 not 0.32

This is a problem on joint probability. Marginal distributions would be used to find the parameters.

PMF = probability mass function

X = the number of hoses being used on the self-service island at a particular time

Y = the number of hoses on the full-service island at that time.

The variable values for (x,y): 0,1, and 2.

(a) To get the conditional PMF of y, we would first calculate the marginal PMF of X and marginal PMF of Y.

Marginal PMF of X= addition of values of x at 0 when y= 0, 1, 2

P (X = 0)= 0.1 + 0.03 + 0.01 = 0.14

Marginal PMF of X= addition of values of x at 1 when y= 0, 1, 2

P (X = 1) = 0.08 + 0.2 + 0.06 = 0.34

Marginal PMF of X= addition of values of x at 2 when y= 0, 1, 2

P (X = 2) = 0.05 + 0.14 + 0.33 = 0.52

Marginal PMF of Y= addition of values of y at 0 when x= 0, 1, 2

P (Y = 0)= 0.1 + 0.08 + 0.05= 0.23

Marginal PMF of Y= addition of values of y at 1 when x= 0, 1, 2

P (Y = 1) = 0.03 + 0.2 + 0.14 = 0.37

Marginal PMF of Y= addition of values of y at 2 when x= 0, 1, 2

P (Y = 2) = 0.01 + 0.06 + 0.33 = 0.40

Since X = 1, the conditional PMF of Y would be:

pY|X(0|1): Y= 0, X = 1

pY|X(0|1) = P(x,y) / P(X= 1) = P(1,0) from the table/ P(X= 1)

pY|X(0|1) = 0.08/0.34 = 0.2353

pY|X(1|1): Y= 1, X = 1

pY|X(1|1) = P(x,y) / P(X= 1) = P(1,1) from the table/ P(X= 1)

pY|X(0|1) = 0.2/0.34 =0.5882

pY|X(2|1): Y= 2, X = 1

pY|X(2|1) = P(x,y) / P(X= 1) = P(2,1) from the table/ P(X= 1)

pY|X(2|1) = 0.06/0.34 =0.1765

b) the conditional PMF of Y would be:

pY|X(?|2): Y= unknown, X = 2

pY|X(Y|2) = P(x,y) / P(X= 1) = P(2,Y) from the table/ P(X= 2)

pY|X(Y|2) =

[P(2,0) + P(2,1) + P(2,2)] / P(X= 2)

= (0.05+0.14+0.33)/ 0.52 = 0.52/0.52 = 1

=1.0000 (4 d.p)

Therefore the conditional pmf of the number of hoses in use on the full-service island is 1.0000

c) pY|X(less than or equal to 1|2): Y = 0,1

= [P(2,0)+ P(2,1)]/ P(X= 2)

= (0.05+0.14)/0.52 = 0.3654

The conditional probability is 0.3846

d) pY|X(2|?): X= unknown, Y = 2

pY|X(2|X) = P(x,y) / P(Y= 2) = P(X,2) from the table/ P(Y= 2)

pY|X(2|X) =

[P(0,2) + P(1,2) + P(2,2)] / P(Y= 2)

pY|X(2|X) = (0.01+0.06+0.33)/0.40

pY|X(2|X) = 1

=1.0000 (4 d.p)

Since the two hoses are in use at the full-service island, the conditional pmf of the number of hoses in use at the self service island is 1.0000

6 on both sides and twelve on the top and the bottom

step-by-step explanation:

d.) yes, because the volume of the cone is greater than the volume of the sphere

step-by-step explanation: