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

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!