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 Y�i.e., pY|X(0|1), pY|X(1|1), pY|X(2|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?
(c) Use the result of part (b) to calculate the conditional probability P(Y ? 1 | X = 2).
(c) Use the result of part (b) to calculate the conditional probability P(Y ? 1 | X = 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} & P(y & {0.0962} & {0.2692} & {0.6346} \ \end{array}`

(c)

`P(Y \le 1 | X = 2) = 0.3654`

Explanation:

Given

`\begin{array}{cccc}{P(x,y)} & {y} & { } & { } &{x } & {0} & {1} & {2} & {0} & {0.10} & {0.03} & {0.01} & {1} & {0.08} & {0.20} & {0.06} & {2} & {0.05} & {0.14} & {0.33} \ \end{array}`

Solving (a): Given x = 1; Find PMF(Y)

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

`\begin{array}{cccc}{P(x,y)} & {y} & { } & { } &{x } & {0} & {1} & {2} & {1} & {0.08} & {0.20} & {0.06} \ \end{array}`

First, calculate P(x=1)

This implies that, we add up the rows where x = 1

So, we have:

`P(x = 1) = 0.08 + 0.20 + 0.06`

`P(x = 1) = 0.34`

The pmf of y is then calculated using:

`PY|X_((yi))=P(Y=y_i|x)=(P(Y=y_i\ and\ x))/(P(x))`

When x = 1

`P(x = 1) = 0.34`

So, we have:

`P(Y=y_i|x)=(P(Y=y_i\ and\ x))/(0.34)`

For i = 0 to 2, we have:

`i = 0`

`P(Y=y_0|x)=(P(Y=0\ and\ x))/(0.34)`

`P(Y=y_0|x)=(0.08)/(0.34)`

`P(Y=y_0|x)=0.2353`

`i = 1`

`P(Y=y_1|x)=(P(Y=1\ and\ x))/(0.34)`

`P(Y=y_1|x)=(0.20)/(0.34)`

`P(Y=y_1|x)=0.5882`

`i = 2`

`P(Y=y_2|x)=(P(Y=2\ and\ x))/(0.34)`

`P(Y=y_2|x)=(0.06)/(0.34)`

`P(Y=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): Given x = 2; Find PMF(Y)

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

`\begin{array}{cccc}{P(x,y)} & {y} & { } & { } &{x } & {0} & {1} & {2} & {2} & {0.05} & {0.14} & {0.33} \ \end{array}`

First, calculate P(x=2)

This implies that, we add up the rows where x = 2

So, we have:

`P(x =2) = 0.05 + 0.14 + 0.33`

`P(x =2) = 0.52`

The pmf of y is then calculated using:

`PY|X_((yi))=P(Y=y_i|x)=(P(Y=y_i\ and\ x))/(P(x))`

When x = 2

`P(x =2) = 0.52`

So, we have:

`P(Y=y_i|x)=(P(Y=y_i\ and\ x))/(0.52)`

For i = 0 to 2, we have:

`i = 0`

`P(Y=y_0|x)=(P(Y=y_0\ and\ x))/(0.52)`

`P(Y=y_0|x)=(0.05)/(0.52)`

`P(Y=y_0|x)=0.0962`

`i = 1`

`P(Y=y_1|x)=(P(Y=y_1\ and\ x))/(0.52)`

`P(Y=y_1|x)=(0.14)/(0.52)`

`P(Y=y_1|x)=0.2692`

`i = 2`

`P(Y=y_2|x)=(P(Y=y_2\ and\ x))/(0.52)`

`P(Y=y_2|x)=(0.33)/(0.52)`

`P(Y=y_2|x)=0.6346`

So, the PMF of y given that x = 2 is:

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

Solving (c):
`P(Y \le 1 | X = 2)`

This means that, we consider the values of Y for which
`Y \le 1` is true

These values are 0 and 1

So, we have:

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

Hence;

`P(Y \le 1 | X = 2) = 0.0962 + 0.2692`

`P(Y \le 1 | X = 2) = 0.3654`