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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
x 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 Yi.e., pY|X(0|1), pY|X(1|1), pY|X(2|1). (Round your answers to four decimal places.)
y 0 1 2
pY|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 2
pY|X(y|2)

Sagot :

MrRoyalidn

Answer:

(a)

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]

(b)

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]

Step-by-step explanation:

Given

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {0} &{0.1}&{0.03} & {0.01} & {0.14} & {1} &{0.08}&{0.20} & {0.06} &{0.34}& {2} &{0.05}&{0.14} & {0.33} &{0.52 } \ \end{array}[/tex]

Solving (a): Find PMF of y if x = 1

This implies that, we consider the dataset of the row where x = 1 i.e.

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {1} &{0.08}&{0.20} & {0.06} &{0.34} \ \end{array}[/tex]

The PMF of y given x is calculated using:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{P(x)}[/tex]

When x = 1

[tex]P(x) = 0.34[/tex] --- the sum of the rows

So, we have:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.34}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=0|x)=\frac{0.08}{0.34}[/tex]

[tex]P(y=0|x)=0.2353[/tex]

[tex]i =1[/tex]

[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=1|x)=\frac{0.20}{0.34}[/tex]

[tex]P(y=1|x)=0.5882[/tex]

[tex]i = 2[/tex]

[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 1)}{0.34}[/tex]

[tex]P(y=2|x)=\frac{0.06}{0.34}[/tex]

[tex]P(y=2|x)=0.1765[/tex]

So, the PMF of y given that x = 1 is:

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|1)} &{0.2353}&{0.5882} & {0.1765}\ \end{array}[/tex]

Solving (b): The interpretation of (b) is to find the PMF of y if x = 2

This implies that, we consider the dataset of the row where x = 2 i.e.

[tex]\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {2} &{0.05}&{0.14} & {0.33} &{0.52 }\ \end{array}[/tex]

When [tex]x = 2[/tex]

[tex]P(x) = 0.52[/tex]

So, we have:

[tex]Py|x(yi)=P(y=y_i|x)=\frac{P(y=x_i\ \&\ x)}{0.52}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(y=0|x)=\frac{P(y=0\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=0|x)=\frac{0.05}{0.52}[/tex]

[tex]P(y=0|x)=0.0962\\[/tex]

[tex]i =1[/tex]

[tex]P(y=1|x)=\frac{P(y=1\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=1|x)=\frac{0.14}{0.52}[/tex]

[tex]P(y=1|x)=0.2692[/tex]

[tex]i = 2[/tex]

[tex]P(y=2|x)=\frac{P(y=2\ \&\ x = 2)}{0.52}[/tex]

[tex]P(y=2|x)=\frac{0.33}{0.52}[/tex]

[tex]P(y=2|x)=0.6346[/tex]

[tex]P(y=0|x)=0.0962\\[/tex]

[tex]P(y=1|x)=0.2692[/tex]

[tex]P(y=2|x)=0.6346[/tex]

So, the PMF of y given that x = 1 is:

[tex]\begin{array}{cccc} {y} & {0} & {1} & {2} & {p(y|2)} &{0.0962}&{0.2692} & {0.6346}\ \end{array}[/tex]

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