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 0.05 0.02 1 0.07 0.20 0.08 2 0.06 0.14 0.28 (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). (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) (c) Use the result of part (b) to calculate the conditional probability P(Y ≤ 1 | X = 2). (Round your answer to four decimal places.) 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? (Round your answers to four decimal places.) x 0 1 2 pX|Y(x|2)

Answer 1

Solution

Given joint pmf of X and Y in a more comprehensive table

x	y			Total
	0	1	2
0	0.10	0.07	0.06	0.23
1	0.05	0.20	0.14	0.39
2	0.02	0.08	0.28	0.38
Total	0.17	0.35	0.48	1.00

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)…….................................................….(1)

Now, to work out the solution,

Part (a)

Given that X = 1, the conditional pmf of Y

= P(Y = y/X = 1)

= P(Y = y, X = 1)/P(X = 1) [vide (1)]

PY(0/X = 1)	PY(1/X = 1)	PY(2/X = 1)	Total
0.1282	0.5128	0.3590	1

Answer 1

Part (b)

Given that two hoses are in use at the self-service island, the conditional pmf of the number of hoses in use on the full-service island

= The conditional pmf of Y given that X = 2

= P(Y = y/X = 2)

= P(Y = y, X = 2)/P(X = 2) [vide (1)]

PY(0/X = 2)	PY(1/X = 2)	PY(2/X = 2)	Total
0.0526	0.2105	0.7368	1

Answer 2

Part (c)

The conditional probability P(Y ≤ 1|X = 2)

= P_Y(0/X = 2) + P_Y(1/X = 2)

= 0.2631 [vide Answer of Part (b)] Answer 3

Part (d)

Given that two hoses are in use at the full-service island, the conditional pmf of the number in use at the self-service island

= P(X = x/Y = 2)

= P(X = x, Y = 2)/P(Y = 2) [vide (1)]

PX(X = 0/Y = 2)	0.1250
PX(X = 1/Y = 2)	0.2917
PX(X = 2/Y = 2)	0.5833
Total	1.0000

Answer 4

DONE