# Another probability problem

#### boks

Joined Oct 10, 2008
218
The amount of kerosene in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x<=y, and assume that the joint density function of these variables is

f(x,y) = 2, 0<x<y<1
f(x,y) = 0, elsewhere

a) Determine if X and Y are independent.
b) Find P(1/4 < X < 1/2 | Y = 3/4).

a)

$g(x) = \int ^1 _{0}2 dy = 2$
$h(y) = \int ^y _{0}2 dx = 2y$

$g(x)h(y) = f(x,y)$, so they are independent.

b)

$f(x|y) = \frac{f(x,y)}{h(y)} = \frac{1}{y}$

#### steveb

Joined Jul 3, 2008
2,431
The amount of kerosene in a tank at the beginning of any day is a random amount Y from which a random amount X is sold during that day. Suppose that the tank is not resupplied during the day so that x<=y, and assume that the joint density function of these variables is

f(x,y) = 2, 0<x<y<1
f(x,y) = 0, elsewhere

a) Determine if X and Y are independent.
b) Find P(1/4 < X < 1/2 | Y = 3/4).

a)

$g(x) = \int ^1 _{0}2 dy = 2$
$h(y) = \int ^y _{0}2 dx = 2y$

$g(x)h(y) = f(x,y)$, so they are independent.

b)

$f(x|y) = \frac{f(x,y)}{h(y)} = \frac{1}{y}$
Your answer to part A has a logical error in it. Clearly the amounts must be dependent since you can't sell more than you have.

An example of independent amounts would be.

1 the amount in the tank at the beginning of the day
2 the amount ordered by customers

assuming that people requesting to purchase have no way of knowing what is in the tank, or if other customers have been turned away.