A service station has both self-service and full-service islands. On each island

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 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 and Y appears in the accompanying tabulation. (Give answers accurate to 3 decimal places.) p(x,y) 0 0.10 0.08 0.04 0.04 0.18 0.16 0 0.02 0.08 0.30 (a) Given that X-1, determine the conditional pmf of Y-ie, 011), Pr x (111), p:1x (211). Y P(Yx-1) 0 0.08 1 0.18 2 0.08 (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? 0 0.04 1 0.16 2 0.30 = 2). (c) Use the result of part (b) to calculate the conditional probability P(Y 0.2 1 (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? p(XIY= 2) 0 0.02 1 0.08 2 0.30

Explanation / Answer

a) Given X=1, we get here:

P(X =1 ) = 0.08 + 0.18 + 0.08 = 0.34

Therefore the conditional probability here is obtained as:

P(Y = 0 | X = 1) = 0.08 / 0.34 = 0.235

P(Y = 1 | X = 1) = 0.18 / 0.34 = 0.529

P(Y = 2 | X = 1) = 0.08 / 0.34 = 0.235
This is the required conditional PDF for Y given X = 1

b) Now here we have: P(X = 2) = 0.04 + 0.16 + 0.30 = 0.50

The conditional probability of Y given X here is computed as:

P(Y = 0 | X = 2) = 0.04 / 0.50 = 0.080

P(Y = 1 | X = 2) = 0.16 / 0.50 = 0.320

P(Y = 2 | X = 2) = 0.30 / 0.50 = 0.060

This is the required conditional PDF for Y given X = 2

c) The required probability here is computed as:

P(Y <=1 | X = 2) = 0.08 + 0.32 = 0.40

Therefore 0.400 is the required probability here.

d) Here, we have P(Y = 2) = 0.02 + 0.08 + 0.30 = 0.40

The conditional probability of X given Y here is computed as:

P(X = 0 | Y = 0) = 0.02/0.40 = 0.050

P(X = 1 | Y = 0) = 0.08/0.40 = 0.200

P(X = 2 | Y = 0) = 0.30/0.40 = 0.750

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