Suppose we have $n$ white and $m$ black balls in a urn. First, randomly withdraw two balls, what is the probability (Call it $P_1$) that they are the same color? Now, suppose a ball is randomly withdrawn and then replaced before second one is drawn, what is the probability (Call it $P_2$) that withdrawn balls are same color? Finally prove that $P_2 > P_1$.
try
For the first situation sample space size is ${m + n \choose 2 }$. Now, in how many ways can we withdraw balls the same color? If both are white, then can do this in ${n \choose 2}$ ways and if both are black can do in ${m \choose 2}$. Thus
$$ P_1 = \frac{ {m \choose 2 } + {n \choose 2} }{ {m+n \choose 2 } } $$
Now, for second situation, two cases. If the first ball drawn is white, then the probability this happens is ${n \choose 1 } / {m+n \choose 1 } = \frac{n}{m+n} $ . For the seecond ball we want it to be white so this can be done in ${n-1 \choose 1 } / {m+n-1 \choose 1 } = \frac{n-1}{m+n-1} $ so for this case we have $\frac{n(n-1) }{(m+n)(m+n-1)}$. Similarly if the first ball drawn is black we obtain probability $ \frac{m(m-1) }{(m+n)(m+n-1)}$.Thus,
$$ P_2 = \frac{ m(m-1) + n(n-1) }{(m+n)(m+n-1) } $$
But, Im stuck in trying to prove $P_2 > P_1$. Is my approach correct?
Best Answer
When working without replacement, we can either select $2$ white balls with probability
$$\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}$$
or $2$ black balls with probability
$$\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
giving $$P_1=\frac{n}{n+m}\cdot\frac{n-1}{n+m-1}+\frac{m}{n+m}\cdot\frac{m-1}{n+m-1}$$
which is equivalent to what you have done.
When working with replacement, the probability does not change after the first draw giving
$$P_2=\frac{n}{n+m}\cdot\frac{n}{n+m}+\frac{m}{n+m}\cdot\frac{m}{n+m}$$
It suffices to show that for two positive integers $$\frac{x}{x+y}\gt\frac{x-1}{x+y-1}$$
Can you go from here?