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
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)

Answer 1

(a)

`\begin{array}{cccc} {y} & {0} & {1} & {2} & 1) &{0.2353}&{0.5882} & {0.1765}\ \end{array}`

(b)

`\begin{array}{cccc} {y} & {0} & {1} & {2} & 2) &{0.0962}&{0.2692} & {0.6346}\ \end{array}`

Explanation:

Given

`\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}`

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

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

`\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {1} &{0.08}&{0.20} & {0.06} &{0.34} \ \end{array}`

The PMF of y given x is calculated using:

`Py|x(yi)=P(y=y_i|x)=(P(y=x_i\ \&\ x))/(P(x))`

When x = 1

`P(x) = 0.34` --- the sum of the rows

So, we have:

`Py|x(yi)=P(y=y_i|x)=(P(y=x_i\ \&\ x))/(0.34)`

For i = 0 to 2, we have:

`i = 0`

`P(y=0|x)=(P(y=0\ \&\ x = 1))/(0.34)`

`P(y=0|x)=(0.08)/(0.34)`

`P(y=0|x)=0.2353`

`i =1`

`P(y=1|x)=(P(y=1\ \&\ x = 1))/(0.34)`

`P(y=1|x)=(0.20)/(0.34)`

`P(y=1|x)=0.5882`

`i = 2`

`P(y=2|x)=(P(y=2\ \&\ x = 1))/(0.34)`

`P(y=2|x)=(0.06)/(0.34)`

`P(y=2|x)=0.1765`

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

`\begin{array}{cccc} {y} & {0} & {1} & {2} & p(y &{0.2353}&{0.5882} & {0.1765}\ \end{array}`

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.

`\begin{array}{ccccc}{} & {y} & { } & { }& { } & {x} & {0} & {1} & {2} & {total} & {2} &{0.05}&{0.14} & {0.33} &{0.52 }\ \end{array}`

When
`x = 2`

`P(x) = 0.52`

So, we have:

`Py|x(yi)=P(y=y_i|x)=(P(y=x_i\ \&\ x))/(0.52)`

For i = 0 to 2, we have:

`i = 0`

`P(y=0|x)=(P(y=0\ \&\ x = 2))/(0.52)`

`P(y=0|x)=(0.05)/(0.52)`

`P(y=0|x)=0.0962\\`

`i =1`

`P(y=1|x)=(P(y=1\ \&\ x = 2))/(0.52)`

`P(y=1|x)=(0.14)/(0.52)`

`P(y=1|x)=0.2692`

`i = 2`

`P(y=2|x)=(P(y=2\ \&\ x = 2))/(0.52)`

`P(y=2|x)=(0.33)/(0.52)`

`P(y=2|x)=0.6346`

`P(y=0|x)=0.0962\\`

`P(y=1|x)=0.2692`

`P(y=2|x)=0.6346`

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

`\begin{array}{cccc} {y} & {0} & {1} & {2} & 2) &{0.0962}&{0.2692} & {0.6346}\ \end{array}`