# Discrete probability distribution

## Homework Statement

The problem is as shown in the attatchment.

## Homework Equations

The relevant equations are also given in the attatchment.

## The Attempt at a Solution

My problem is how to adapt the given formula in order to find the sum of the function k(40-r)

Do i use the formula for 1 to 40, then 1 to 20 and subtract?

Even just a hint in the right direction would be useful. This sorta stuff has never been my strong point.

#### Attachments

• q1.PNG
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Think i might have got it.

k(1$$\rightarrow$$40$$\sum(40-r)$$ - 1$$\rightarrow$$19$$\sum(40-r)$$) + 20k = 1

k(40*40 - $$\frac{40(41)}{2}$$ - (19*20 - $$\frac{19(20)}{2}$$)) = 1

230k =1
k = 1/230

EDIT sorry its a bit messy, dont know how to notate the limits for the summation properly.

Last edited:
How did

$$\sum_{r=1}^{19}(40-r) = (19)(20) - \frac{(19)(20)}{2}$$

The part I don't understand is the (19)(20). And what happened to the 20k? Should it be 20k?

tiny-tim
Homework Helper
Hi ineedmunchies!

∑(40 - r) from 20 to 40 is simply:

20 + 19 + … + 1 + 0.

Just sum that!

How did

$$\sum_{r=1}^{19}(40-r) = (19)(20) - \frac{(19)(20)}{2}$$

The part I don't understand is the (19)(20). And what happened to the 20k? Should it be 20k?

I'm pretty sure that should be 19*40 and the 20k should be taken into account too. But as tim said, I've made it too complicated.

Tim did make a good observation. Make sure, however, that you don't use 20k. What should it be instead?

So can anyone help on how to get the expected value of R?

I think you multiply the two functions by r and then work out the sum over the ranges. Not entirely sure.

Oh and does anybody agree with my value of 1/230 for k??

EDIT: sorry didn't see your post tedjn.

I think you have the right idea, but how many numbers are between 0 and 19 inclusive?

ahhh silly little mistake again. twenty, ok getting there slowly.

1 = 20*20k + (20*21/2)k = 610k
k=1/610 hows that look?

Yeah, that looks good.

tiny-tim