Respuesta :
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!
(a)
Case 1: both balls are white.
At the beginning we have [tex]b+w[/tex] balls. We want to pick a white one, so we have a probability of [tex]\frac{w}{b+w}[/tex] of picking a white one.
If this happens, we're left with [tex]w-1[/tex] white balls and still [tex]b[/tex] black balls, for a total of [tex]b+w-1[/tex] balls. So, now, the probability of picking a white ball is
[tex]\dfrac{w-1}{b+w-1}[/tex]
The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability
[tex]\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}[/tex]
Case 2: both balls are black
The exact same logic leads to a probability of
[tex]\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}[/tex]
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
[tex]\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)}[/tex]
(b)
Case 1: both balls are white.
In this case, nothing changes between the two picks. So, you have a probability of [tex]\frac{w}{b+w}[/tex] 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 [tex]\frac{b}{b+w}[/tex] of picking a black ball with both picks.
This leads to an overall probability of
[tex]\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}[/tex]
Of picking two balls of the same colour.
(c)
We want to prove that
[tex]\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}[/tex]
Expading all squares and products, this translates to
[tex]\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}[/tex]
As you can see, this inequality comes in the form
[tex]\dfrac{x}{y}\geq \dfrac{x-k}{y-k}[/tex]
With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:
[tex]\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y[/tex]
And this is our case, because in our case we have
- [tex]x=b^2+w^2[/tex]
- [tex]y=b^2+w^2+2bw[/tex] so, y has an extra piece and it is larger
- [tex]k=b+w[/tex] which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2
