91Ó°ÊÓ

An urn contains $n$white and $m$black balls, where$n$and$m$ are positive numbers.

An urn contains $n$white and $m$black balls, where $n$and$m$are positive numbers. The probability is equal to the ratio of withdrawal of a ball of the same color and every possible withdrawal.

Assume that every ball is equally likely to be withdrawn. Therefore, probability$P$can be calculated:

$$\begin{gathered} P=\frac{\left(\begin{array}{c}m\\ 2\end{array}\right)+\left(\begin{array}{c}n\\ 2\end{array}\right)}{\left(\begin{array}{c}m+n\\ 2\end{array}\right) \\ }=\frac{\frac{m^2-m+n^2-n}{2}}{\frac{(m+n)(m+n-1)}{2}}=\frac{m^2+n^2-m-n}{m^2+n^2+2mn-m-n}. \end{gathered}$$

If the ball is randomly withdrawn and replaced before the second one is drawn the probability is obviously larger since there are again $n$white (or $m$black) balls instead of $n-1(m-1)$The probability is precise:

$P=\frac{m^2+n^2}{(m+n{)}^2}$

Now we are left with proving that the probability in part$\left(b\right)$is larger than the probability in part$\left(a\right)$. Since we have them explicitly we can show that easily:

$\frac{m^2+n^2}{(m+n{)}^2}>\frac{m^2+n^2-m-n}{m^2+n^2+2mn-m-n}$

$⇔\left(m^2+n^2\right)(m+n{)}^2-(m+n)\left(m^2+n^2\right)>\left(m^2+n^2\right)(m+n{)}^2-(m+n{)}^3$

After canceling the inequality is equivalent to:

$(m+n{)}^3>(m+n)\left(m^2+n^2\right)⇔(m+n{)}^2>m^2+n^2$

which is trivial. Therefore, we are done.