Q. 2.29 - A First Course in Probability

Question

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

$\left(a\right)$ If two balls are randomly withdrawn, what is the probability that they are the same color?

$\left(b\right)$ If a ball is randomly withdrawn and then replaced before the second one is drawn, what is the probability that the withdrawn balls are the same color?

$\left(c\right)$ Show that the probability in part $\left(b\right)$ is always larger than the one in part $\left(a\right)$.

Step-by-Step Solution

Answer

The probabilities are $\frac{m^2+n^2-m-n}{(m+n{)}^2-m-n}$and $\frac{m^2+n^2}{(m+n{)}^2}$respectively, while the inequality follows trivially after simplifying.

Step 1 Given Information.

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

Step 2 Part (a) Explanation.

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Assume that every ball is equally likely to be withdrawn.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:

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

Step 3 Part (b) Explanation.

If the ball is randomly withdrawn and replaced before the second one is drawn the probability is obviously larger since there are again $n$white (or $m$black) balls instead of $n-1(m-1)$ The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

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

Step 4 Part (c) Explanation.

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:

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

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

After canceling the inequality is equivalent to:

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

which is trivial. Therefore, we are done.