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 0.05 0.01 1 0.10 0.06 0.06 0.20 0.07 2 0.14 0.31 (a) Given that X = 1, determine the conditional pmf of Y-i.e., Py\x(011), Pyx(1/1), Py\x{2/1). (Round your answers to four decimal places.) 0 1 2 y Prix(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.) у Pyx(12) 2 + (c) Use the result of part (b) to calculate the conditional probability PY $1|X = 2). (Round your answer to four decimal places.) PYS1X = 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.) 0 2 Pxy(x2)

Answer 1

The joint pmf is given by:

			Y
	f(x,y)	0	1	2	Total
	0	0.1	0.05	0.01	0.16
X	1	0.06	0.2	0.07	0.33
	2	0.06	0.14	0.31	0.51
	Total	0.22	0.39	0.39	1

• Here, we are to find:
  $P_{Y|X}(y|1)$
  $=\frac{P_{Y,X}(y,1)}{P(X=1)}$
  Now,
  
  P(X = 1) = 0.33
  
  
  We compute the following table for ease of calculation:

  Y	P(X=1,Y=y)	P(X=1)	P(Y=y|X=1)
  0	0.06	0.33	0.181818182
  1	0.2	0.33	0.606060606
  2	0.07	0.33	0.212121212
  Total	-	-	1

  
  Hence, the conditional distribution is given by:

  Y	0	1	2
  $P_{Y|X}(y|1)$	0.1818	0.6061	0.2121

  
  
• Here, we are to find:
  $P_{Y|X}(y|2)$
  $=\frac{P_{Y,X}(y,2)}{P(X=2)}$
  Now,
  $P(X=2)=0.51$
  
  We compute the following table for ease of calculation:

  Y	P(X=2,Y=y)	P(X=2)	P(Y=y|X=2)
  0	0.06	0.51	0.117647059
  1	0.14	0.51	0.274509804
  2	0.31	0.51	0.607843137
  Total	-	-	1

  
  Hence, the conditional distribution is given by:

  Y	0	1	2
  $P_{Y|X}(y|2)$	0.1176	0.2745	0.6078

  
  
• Here, we are to find:
  $P_{Y|X}(Y \leq 1|2)$
  
  $=0.1176+0.2745$
  $=0.3921$
  
• Here, we are to find:
  $P_{X|Y}(x|2)$
  $=\frac{P_{Y,X}(2,x)}{P(Y=2)}$
  Now,
  $P(Y=2)=0.39$
  
  We compute the following table for ease of calculation:

  X	P(X=x,Y=2)	P(Y=2)	P(X=x|Y=2)
  0	0.01	0.39	0.025641026
  1	0.07	0.39	0.179487179
  2	0.31	0.39	0.794871795
  Total	-	-	1

  
  Hence, the conditional distribution is given by:

  X	0	1	2
  $P_{X|Y}(x|2)$	0.0256	0.1795	0.7949

  
  

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