DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!
<h2>(a)</h2>
Case 1: both balls are white.
At the beginning we have
balls. We want to pick a white one, so we have a probability of
of picking a white one.
If this happens, we're left with
white balls and still
black balls, for a total of
balls. So, now, the probability of picking a white ball is
![\dfrac{w-1}{b+w-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bw-1%7D%7Bb%2Bw-1%7D)
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
![\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bw%7D%7Bb%2Bw%7D%5Ccdot%20%5Cdfrac%7Bw-1%7D%7Bb%2Bw-1%7D%3D%5Cdfrac%7Bw%28w-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D)
Case 2: both balls are black
The exact same logic leads to a probability of
![\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bb%7D%7Bb%2Bw%7D%5Ccdot%20%5Cdfrac%7Bb-1%7D%7Bb%2Bw-1%7D%3D%5Cdfrac%7Bb%28b-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D)
These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is
![\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bw%28w-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D%2B%5Cdfrac%7Bb%28b-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D%3D%5Cdfrac%7Bw%28w-1%29%2Bb%28b-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D)
<h2>(b)</h2>
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of
of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability
of picking a black ball with both picks.
This leads to an overall probability of
![\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}](https://tex.z-dn.net/?f=%5Cleft%28%5Cdfrac%7Bw%7D%7Bb%2Bw%7D%5Cright%29%5E2%2B%5Cleft%28%5Cdfrac%7Bb%7D%7Bb%2Bw%7D%5Cright%29%5E2%20%3D%20%5Cdfrac%7Bw%5E2%2Bb%5E2%7D%7B%28b%2Bw%29%5E2%7D)
Of picking two balls of the same colour.
<h2>(c)</h2>
We want to prove that
![\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}](https://tex.z-dn.net/?f=%5Cdfrac%7Bw%5E2%2Bb%5E2%7D%7B%28b%2Bw%29%5E2%7D%5Cgeq%20%5Cdfrac%7Bw%28w-1%29%2Bb%28b-1%29%7D%7B%28b%2Bw%29%28b%2Bw-1%29%7D)
Expading all squares and products, this translates to
![\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}](https://tex.z-dn.net/?f=%5Cdfrac%7Bw%5E2%2Bb%5E2%7D%7Bb%5E2%2B2bw%2Bw%5E2%7D%5Cgeq%20%5Cdfrac%7Bw%5E2%2Bb%5E2-b-w%7D%7Bb%5E2%2B2bw%2Bw%5E2-b-w%7D)
As you can see, this inequality comes in the form
![\dfrac{x}{y}\geq \dfrac{x-k}{y-k}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%7D%7By%7D%5Cgeq%20%5Cdfrac%7Bx-k%7D%7By-k%7D)
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
![\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%7D%7By%7D%5Cgeq%20%5Cdfrac%7Bx-k%7D%7By-k%7D%20%5Ciff%20xy-kx%20%5Cgeq%20xy-ky%20%5Ciff%20-kx%5Cgeq%20-ky%20%5Ciff%20x%5Cleq%20y)
And this is our case, because in our case we have
![x=b^2+w^2](https://tex.z-dn.net/?f=x%3Db%5E2%2Bw%5E2)
so, y has an extra piece and it is larger
which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2