# Thread: [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.

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

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# A candy company distribute boxes of chocolates with a mixture of creams, toffees and cordials. Suppose that the weight of each box is 1kg but the individual weights of chocolates creams toffees and cordials vary from box to box. For a random selected box,

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