The probability of drawing both without replacing either is :
14/41 * 12/40 = 7/41 * 3/5 = 21/205
The probability of drawing both WITH replacement is:
14/41 * 12/41 = 168/ 1681

This distribution has expectation
![E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\int_1^\infty\frac3{x^3}\,\mathrm dx=\frac32](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac3%7Bx%5E3%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac32)
a. The probability that
falls below the average/expectation is

b. Denote by
the largest of the three claims
. Then the density of this maximum order statistic is

where
is the distribution function for
. This is given by

So we have

and the expectation is
![E[X_{(3)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(3)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^3}\left(1-\frac1{x^3}\right)^2\,\mathrm dx=\frac{81}{40}=\boxed{2.025}](https://tex.z-dn.net/?f=E%5BX_%7B%283%29%7D%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_%7BX_%7B%283%29%7D%7D%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac9%7Bx%5E3%7D%5Cleft%281-%5Cfrac1%7Bx%5E3%7D%5Cright%29%5E2%5C%2C%5Cmathrm%20dx%3D%5Cfrac%7B81%7D%7B40%7D%3D%5Cboxed%7B2.025%7D)
c. Denote by
the smallest of the three claims.
has density

so the expectation is
![E[X_{(1)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(1)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^9}\,\mathrm dx=\frac98=\boxed{1.125}](https://tex.z-dn.net/?f=E%5BX_%7B%281%29%7D%5D%3D%5Cdisplaystyle%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20xf_%7BX_%7B%281%29%7D%7D%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cint_1%5E%5Cinfty%5Cfrac9%7Bx%5E9%7D%5C%2C%5Cmathrm%20dx%3D%5Cfrac98%3D%5Cboxed%7B1.125%7D)
The answer is C!!!!!!!!!!!
Answer: Ok, so this question is confusing to me as well, and I have this same question on school. So when I get the answer, I will try to explain it to you the best I can. :) God Bless. Stay :D
Step-by-step explanation: