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.
p(x, y) 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).
(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?

Answer 1

Answer:

(a): The conditional pmf of Y when X = 1

$p_{Y|X}(0|1) = 0.2353$

$p_{Y|X}(1|1) = 0.5882$

$p_{Y|X}(2|1) = 0.1765$

(b): The conditional pmf of Y when X = 2

$p_{Y|X}(0|2) = 0.0962$

$p_{Y|X}(1|2) = 0.2692$

$p_{Y|X}(2|2) = 0.6346$

(c): From (b) calculate P(Y<=1 | X =2)

$P(Y\le1 | X =2) = 0.3654$

(d): The conditional pmf of X when Y = 2

$p_{X|Y}(0|2) = 0.025$

$p_{X|Y}(1|2) = 0.150$

$p_{X|Y}(2|2) = 0.825$

Step-by-step explanation:

Given

The above table

Solving (a): The conditional pmf of Y when X = 1

This implies that we calculate

$p_{Y|X}(0|1), p_{Y|X}(1|1), p_{Y|X}(2|1)$

So, we have:

$p_{Y|X}(0|1) = \frac{p(y = 0\ n\ x = 1)}{p(x = 1)}$

Reading the data from the given table, the equation becomes

$p_{Y|X}(0|1) = \frac{0.08}{0.08+0.20+0.06}$

$p_{Y|X}(0|1) = \frac{0.08}{0.34}$

$p_{Y|X}(0|1) = 0.2353$

Using the format of the above formula for the rest, we have:

$p_{Y|X}(1|1) = \frac{0.20}{0.34}$

$p_{Y|X}(1|1) = 0.5882$

$p_{Y|X}(2|1) = \frac{0.06}{0.34}$

$p_{Y|X}(2|1) = 0.1765$

Solving (b): The conditional pmf of Y when X = 2

This implies that we calculate

$p_{Y|X}(0|2), p_{Y|X}(1|2), p_{Y|X}(2|2)$

So, we have:

$p_{Y|X}(0|2) = \frac{p(y = 0\ n\ x = 2)}{p(x = 2)}$

Reading the data from the given table, the equation becomes

$p_{Y|X}(0|2) = \frac{0.05}{0.05+0.14+0.33}$

$p_{Y|X}(0|2) = \frac{0.05}{0.52}$

$p_{Y|X}(0|2) = 0.0962$

Using the format of the above formula for the rest, we have:

$p_{Y|X}(1|2) = \frac{0.14}{0.52}$

$p_{Y|X}(1|2) = 0.2692$

$p_{Y|X}(2|2) = \frac{0.33}{0.52}$

$p_{Y|X}(2|2) = 0.6346$

Solving (c): From (b) calculate P(Y<=1 | X =2)

To do this, where Y = 0 or 1

So, we have:

$P(Y\le1 | X =2) = P_{Y|X}(0|2) + P_{Y|X}(1|2)$

$P(Y\le1 | X =2) = 0.0962 + 0.2692$

$P(Y\le1 | X =2) = 0.3654$

Solving (d): The conditional pmf of X when Y = 2

This implies that we calculate

$p_{X|Y}(0|2), p_{X|Y}(1|2), p_{X|Y}(2|2)$

So, we have:

$p_{X|Y}(0|2) = \frac{p(x = 0\ n\ y = 2)}{p(y = 2)}$

Reading the data from the given table, the equation becomes

$p_{X|Y}(0|2) = \frac{0.01}{0.01+0.06+0.33}$

$p_{X|Y}(0|2) = \frac{0.01}{0.40}$

$p_{X|Y}(0|2) = 0.025$

Using the format of the above formula for the rest, we have:

$p_{X|Y}(1|2) = \frac{0.06}{0.40}$

$p_{X|Y}(1|2) = 0.150$

$p_{X|Y}(2|2) = \frac{0.33}{0.40}$

$p_{X|Y}(2|2) = 0.825$