Q:

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacement, what is the probability that both balls are the same color? (b) If two balls are drawn with replacement (i.e., One ball is drawn and it’s color recorded and then put back. Then the second ball is drawn.) What is the probability that both balls are the same color. (c) Show that the probability in part (b) is always larger than the one in part (a).

Accepted Solution

A:
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 blackThe 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