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 0.04 0.02
1 0.08 0.20 0.06
2 0.06 0.14 0.30

a. What is P(X = 1 and Y = 1)?
b. Compute P(X ≤ 1 and Y ≤ 1).
c. Give a word description of the event {X ≠ 0 and Y ≠ 0}, and compute the probability of this event.
d. Compute the marginal pmf of X and of Y. Using pX(x), what is P(X ≤ 1)?
e. Are X and Y independent rv’s? Explain.

Answer 1

a. 0.2

b. 0.42

c. 0.7

d. the solution is in the explanation

e. x and y are not independent

Explanation:

a. from the joint probability mass function table,

p(x=1) and p(Y= 1)

= p(1,1) = 0.2

b. prob(0,0)+prob(0,1)+prob(1,0)+prob(1,1)

= 0.10 + 0.04 + 0.08 + 0.20

= 0.42

P(X ≤ 1 and Y ≤ 1) = 0.42

c. prob {X ≠ 0 and Y ≠ 0}

= prob(1,1) + prob(1,2) + prob(2,1) + prob(2,2)

= 0.20 + 0.06 + 0.14 + 0.30

= 0.7

d. we have to calculate the marginal pmf of x and y here.

we have the x values as 0,1,2

prob(x=0) = 0.1 + 0.04 + 0.02

= 0.16

prob(x=1) = 0.08 + 0.2 + 0.06

= 0.34

prob(x=2) = 0.06+0.14+0.3

= 0.50

we have y values as 0,1,2

prob(y=0) = .1+.08+.06

= 0.24

prob(y=1) = .04+.2+.14

= 0.38

prob(y = 2) = 0.02+0.06+0.3

= 0.38

P(X ≤ 1) = prob(x=0)+prob(x=1)

= 0.34+0.16

= 0.50

e. from the joint table we have this,

prob(1,1) = 0.2

prob(x=1) = 0.34

prob(y=1) = 0.38

then prob(x=1)*prob(y=1)

= 0.34*0.38

= 0.1292

therefore prob(1,1) is not equal to prob(x=1)*prob(y=1)

0.2≠0.1292

x and y are not independent