# Math Help - [SOLVED] Joint /Conditional Probability Distributions

1. ## [SOLVED] Joint /Conditional Probability Distributions

A candy company distributes boxes of chococlates with a mixture of creams, toffees, and cordials. Suppose that the weight of each box is 1 kg, but the individual weights of the creams, toffees and cordials vary from box to box. For a randomly selected box, let X and Y represent the weights of the creams and the toffees, respectively, and suppose that the joint density function of these var is

f(x,y) = 24xy , 0<= x<=1, 0 <=y<=1, x+y<=1
0 elsewhere

a) find the prob that in a given box the cordials acount for more than 1/2 of the weight.
b) find the mariginal density of the weight of the creams.
c) Find the prob that the weight of the toffees in a box is less than 1/8 of a kg if it is known that creams constitute 3/4 of the weight.

---------------

a) I integrate 24xy from 0 to 1/2 to dy. and my answer= 3/8
however, the correct ans = 1/16
where have i gone wrong ?

b) g(x) = 12xy^2
but the correct answer is 12x(1-x)^2.
why?

c) P(y<1/8 | x=3/4)
= f(x,y) / g(x)
and i got 2 but the correct answer is 1/4.

2. 3.41 from walpole?

you cannot ignore x+y<1 and you need to draw the region (triangle)
also, a probability cannot exceed 1.

3. Oh yes ! this is qn 3.41

Sorry i don't get what you mean. i understand that prb cannot exceed 1 and that x+y <1
which is given. so what should i do ?

4. since x+y<1, these are DEPENDENT rvs

$P(X+Y\le 1/2)= \int_0^{1/2}\int_0^{1/2-y}24xy dxdy=1/16$

$f_X(x)= \int_0^{1-x}24xy dy=12x(1-x)^2$ where 0<x<1 which is a BETA RV

$f(y|x)= {f(x,y)\over f_X(x)}={2y\over (1-x)^2}$

so $P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4$

you should be happy that I teach out of walpole
in fact, look at page xxiii, I'm the first reviewer.

5. Originally Posted by matheagle
since x+y<1, these are DEPENDENT rvs

$P(X+Y\le 1/2)= \int_0^{1/2}\int_0^{1/2-y}24xy dxdy=1/16$

$f_X(x)= \int_0^{1-x}24xy dy=12x(1-x)^2$ where 0<x<1 which is a BETA RV

$f(y|x)= {f(x,y)\over f_X(x)}={2y\over (1-x)^2}$

so $P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4$

you should be happy that I teach out of walpole
in fact, look at page xxiii, I'm the first reviewer.
but why is it for :
$P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4$

I can't just substitute 1/8 and 3/4 into the equation
${2y\over (1-x)^2}$
which i got 4.

6. Originally Posted by hazel
but why is it for :
$P(Y<1/8|X=3/4)=32\int_0^{1/8}y dy=1/4$

I can't just substitute 1/8 and 3/4 into the equation
${2y\over (1-x)^2}$
which i got 4.
YOU substitute the X=3/4 and you integrate Y from -infinity to 1/8
note that all these densities are 0 when the variables are negative
so you integrate from 0 to 1/8.

IN GENERAL $P(Y<1/8|X=3/4)=\int_{-\infty}^{1/8}f(y|X=3/4) dy$

Just like a marginal density, you integrate it over the appropriate region.

It's 1 am, I'm tired

7. ok, let me read thru again.

thanks.

8. i got 27/512.

could you show the step by step workings for
$
P(Y<1/8|X=3/4)=\int_{-\infty}^{1/8}f(y|X=3/4) dy
$

9. It's very basic calc 1

$f(y|X=x)=2y/(1-x)^2$

So $f(y|X=3/4)=32y$ where 0<y<1/4

AND not for 0<y<1, that's not a valid density

This is why I told you to draw the region.
y>0 and x+y<1, with x=3/4 that means that y<1/4 now.

Thus $P(Y<1/8|X=3/4)=32\int_0^{1/8} y dy =16(1/8)^2=16/64=1/4$