1. ## Probability

hello i'm new to things like these and I need your help guys:

two construction companies make bids of X and Y on a remodeling project.The joint p.d.f of X and Y is uniform on the space 2<x<2.5 2<y<2.3

If X and Y are within 0.1 of each other,the companies will be asked to rebid; otherwise the lower bidder will be awarded the contract.
What is the probability that they will be asked to rebid?

I have no idea can you give me at least a guide or hint?

Thank you

2. Originally Posted by qwerty321
hello i'm new to things like these and I need your help guys:

two construction companies make bids of X and Y on a remodeling project.The joint p.d.f of X and Y is uniform on the space 2<x<2.5 2<y<2.3

If X and Y are within 0.1 of each other,the companies will be asked to rebid; otherwise the lower bidder will be awarded the contract.
What is the probability that they will be asked to rebid?

I have no idea can you give me at least a guide or hint?

Thank you
First draw a diagram. The support is a rectangle defined by 2<x<2.5 and 2<y<2.3. Therefore the joint pdf is f(x, y) = 20/3 (why?).

Now note that if the two bids are within 0.1 of each other, then the region of interest lies between the lines y = x - 0.1 and y = x + 0.1.

The answer can be found using simple geometry - no need for calculus.

3. sry i don't really get it
I tried to integrate 20/3 for y=0.1-x to y=0.1+x and for x=2 to x=2.5 but i do not get the correct answer(which should be 11/30)
Plus how did u get 20/3?

thank you

4. ok i got the 20/3
because they are independant fo the joint pmf is that of x times that of y
but then?

5. Originally Posted by qwerty321
ok i got the 20/3
because they are independant fo the joint pmf is that of x times that of y
but then?
OK, now go back and read my first reply. Draw the two lines. Use geometry to calculate the area of the required region. Multiply that area by 20/3.

By the way, if your calculations lead to the wrong answer, you should post all your calculations for review - perhaps there is a simple error. Using calculus is the long way.

6. i will use integral

is it int (20/30 for y=x-0.1 to y=x+0.1 and for x=?

i cannot find the value for x

7. there is no reason to integrate, just draw and figure out the region of interest.

8. in my case i have to integrate
please i have a test tomorrow and i wanan know how to solve this
thank you very much for your help

9. Originally Posted by qwerty321
in my case i have to integrate
please i have a test tomorrow and i wanan know how to solve this
thank you very much for your help
A test on New Years Day?

Have you studied multi-variable calculus, specifically integration over a region? If you have not, then using a calculus approach is pointless. And if simple geometry can get the answer, why are you insisting on using calculus. For all the help it might be, here:

$\displaystyle \int_{x = 2}^{x = 2.2} \int_{y = 2}^{y = x + 0.1} f(x, y) \, dy \, dx + \int^{x = 2.4}_{x = 2.2} \int^{y = 2.3}_{y = x - 0.1} f(x, y) \, dy \, dx$.

Edit: big mistake here due to mistaking a 2.1 for a 2.2 on my diagram. See my later post.

10. the answer is still wrong though
it should be 11/30

11. sry it worked
the x u wrote should be from 2.2 to 2.5 in the second integral
thank you very much

btw one question
if we integrated first with respect to x, we would also have to slipt the integral in two parts?
thank you mr fantastic

12. YOU must draw the region if you want to integrate
AND once you draw it you might even realize that all you need is the area of that strip.
So, integration isn't even necessary.

13. Originally Posted by qwerty321
sry it worked
the x u wrote should be from 2.2 to 2.5 in the second integral
thank you very much

btw one question
if we integrated first with respect to x, we would also have to slipt the integral in two parts?
thank you mr fantastic
I made a mistake in my reading of my diagram (I have to say my reluctance in using calculus here probably contributed to carelessness on my part). The mistake is NOT the 2.5 (integrating to 2.5 is wrong). Did you draw the diagram? It should be very clear that x = 2.5 is NOT on the boundary of the region.

As it happens, if you integrate first wrt y you have to integrate over three seperate regions. So it's best to integrate first wrt to x. Then:

$\displaystyle \int_{y = 2}^{y = 2.1} \int_{x = 2}^{x = y + 0.1} f(x, y) \, dx \, dy + \int_{y = 2.1}^{y = 2.3} \int_{x = y - 0.1}^{x = y + 0.1} f(x, y) \, dx \, dy$.

But, and this has been said several times, why insist on using calculus to do something that can be done so easily using simple geometry .... (But I suppose if you want to spend 15 minutes in an exam on a question that can be done in 5 minutes, then that's your choice).