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).

[tex]\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}[/tex]

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

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

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

First, calculate P(x=1)

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

So, we have:

[tex]P(x = 1) = 0.08 + 0.20 + 0.06[/tex]

[tex]P(x = 1) = 0.34[/tex]

The pmf of y is then calculated using:

[tex]PY|X_{(yi)}=P(Y=y_i|x)=\frac{P(Y=y_i\ and\ x)}{P(x)}[/tex]

When x = 1

[tex]P(x = 1) = 0.34[/tex]

So, we have:

[tex]P(Y=y_i|x)=\frac{P(Y=y_i\ and\ x)}{0.34}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(Y=y_0|x)=\frac{P(Y=0\ and\ x)}{0.34}[/tex]

[tex]P(Y=y_0|x)=\frac{0.08}{0.34}[/tex]

[tex]P(Y=y_0|x)=0.2353[/tex]

[tex]i = 1[/tex]

[tex]P(Y=y_1|x)=\frac{P(Y=1\ and\ x)}{0.34}[/tex]

[tex]P(Y=y_1|x)=\frac{0.20}{0.34}[/tex]

[tex]P(Y=y_1|x)=0.5882[/tex]

[tex]i = 2[/tex]

[tex]P(Y=y_2|x)=\frac{P(Y=2\ and\ x)}{0.34}[/tex]

[tex]P(Y=y_2|x)=\frac{0.06}{0.34}[/tex]

[tex]P(Y=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): Given x = 2; Find PMF(Y)

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

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

First, calculate P(x=2)

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

So, we have:

[tex]P(x =2) = 0.05 + 0.14 + 0.33[/tex]

[tex]P(x =2) = 0.52[/tex]

The pmf of y is then calculated using:

[tex]PY|X_{(yi)}=P(Y=y_i|x)=\frac{P(Y=y_i\ and\ x)}{P(x)}[/tex]

When x = 2

[tex]P(x =2) = 0.52[/tex]

So, we have:

[tex]P(Y=y_i|x)=\frac{P(Y=y_i\ and\ x)}{0.52}[/tex]

For i = 0 to 2, we have:

[tex]i = 0[/tex]

[tex]P(Y=y_0|x)=\frac{P(Y=y_0\ and\ x)}{0.52}[/tex]

[tex]P(Y=y_0|x)=\frac{0.05}{0.52}[/tex]

[tex]P(Y=y_0|x)=0.0962[/tex]

[tex]i = 1[/tex]

[tex]P(Y=y_1|x)=\frac{P(Y=y_1\ and\ x)}{0.52}[/tex]

[tex]P(Y=y_1|x)=\frac{0.14}{0.52}[/tex]

[tex]P(Y=y_1|x)=0.2692[/tex]

[tex]i = 2[/tex]

[tex]P(Y=y_2|x)=\frac{P(Y=y_2\ and\ x)}{0.52}[/tex]

[tex]P(Y=y_2|x)=\frac{0.33}{0.52}[/tex]

[tex]P(Y=y_2|x)=0.6346[/tex]

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

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

Solving (c): [tex]P(Y \le 1 | X = 2)[/tex]

This means that, we consider the values of Y for which [tex]Y \le 1[/tex] is true

These values are 0 and 1

So, we have:

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

Hence;

[tex]P(Y \le 1 | X = 2) = 0.0962 + 0.2692[/tex]

[tex]P(Y \le 1 | X = 2) = 0.3654[/tex]