# Hard Probability Question

#### WeeG

##### TS Contributor
Hello,

I have another probability question, harder this time (I think).

I am attaching 2 pictures, one of the question and 1 of my attempt to solve it.

Any tips will be appreciated !

#### BGM

##### TS Contributor
For the last step:

$$\int_{e^{10}}^{+\infty} e^{20 - 2\ln r} dr = e^{20} \int_{e^{10}}^{+\infty} (e^{\ln r})^{-2} dr = e^{20} \int_{e^{10}}^{+\infty} \frac {dr} {r^2} = e^{20} \left. \frac {-1} {r} \right|_{r = e^{10}} ^ {r = +\infty} = e^{10}$$

#### WeeG

##### TS Contributor
thank you VERY much, you have no idea how much you helped me...how didn't I see that e^20 should get out ???

As for part "b", I think I solved it, what do you think ?

The profit W=1000-(1000/T)

F(W)=P(W<=w)=P(T<=(1000-w)/1000)=1-e^(20-((1000-w)/1000))

After a derivative, I got:

f(W)=-(1/1000)*e^()