91Ó°ÊÓ

There is a box which contains 100 balls.

The number of red balls in the box is r.

Here, we take that at first we ordered the 100 balls in a list. After that,we can draw the balls in order.

Assuming that randomly ordered the balls in a list after that we can draw the balls in that order.

a. Here at first, we ordered 100 balls in a list. After that, the balls are drawn in a random order in the list.

Total number of balls is 100 and total number of red ball is r.

So, the probability of first ball being red is\(\frac{r}{{100}}\).

Therefore, the probability of the event that the first ball on the list will be red is\(\frac{r}{{100}}\).

b.

Let the total number of balls be denoted by n and the total number of red balls denoted by k, then,

\(\begin{aligned}{l}n = 100\\k = r\end{aligned}\)

\(\begin{aligned}{c}{P_{{n_1} + 1}} = \sum\limits_{{k_1} = 0}^{{n_1}} {{P_{n,k}}\left( {{n_1},{k_1}} \right) \times \frac{{k - {k_1}}}{{n - {n_1}}}} \\ = \sum\limits_{{k_1} = 0}^{{n_1}} {\frac{{\left( {\begin{aligned}{*{20}{c}}k\\{{k_1}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{n - k}\\{{n_1} - {k_1}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}n\\{{n_1}}\end{aligned}} \right)}}} \times \frac{{k - {k_1}}}{{n - {n_1}}}\\ = \sum\limits_{{k_1} = 0}^{{n_1}} {\frac{{\frac{{k\left( {{k_1} - 1} \right)!}}{{{k_1}!\left( {k - 1 - {k_1}} \right)!}}\left( {\begin{aligned}{*{20}{c}}{n - k}\\{{n_1} - {k_1}}\end{aligned}} \right)}}{{\frac{{n\left( {n - 1} \right)!}}{{{n_1}!\left( {n - 1 - {n_1}} \right)!}}}}} \\ = \frac{k}{n}\sum\limits_{{k_1} = 0}^{{n_1}} {\frac{{\left( {\begin{aligned}{*{20}{c}}{k - 1}\\{{k_1}}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{n - k}\\{{n_1} - {k_1}}\end{aligned}} \right)}}{{\left( {\begin{aligned}{*{20}{c}}{n - 1}\\{{n_1}}\end{aligned}} \right)}}} \\ = \frac{k}{n} \times 1\\ = \frac{k}{n}\end{aligned}\)

The probability of the event that the first ball in the list will be red is

\(\begin{aligned}{c}{P_{0 + 1}} &= {P_1}\\ &= \frac{k}{n}\\ &= \frac{r}{{100}}\end{aligned}\)

The probability of the event that the second ball in the list will be red is

\(\begin{aligned}{c}{P_{1 + 1}} = {P_2}\\ &= \frac{k}{n}\\ &= \frac{r}{{100}}\end{aligned}\)

Thus, the probability of the event that the 50th ball in the list will be red is

\(\begin{aligned}{c}{P_{49 + 1}} &= {P_{50}}\\ &= \frac{k}{n}\\ &= \frac{r}{{100}}\end{aligned}\)

So, the probability of the event that the 50th ball in the list will be red is\(\frac{r}{{100}}\).

c. Referring to the above answer of Exercise 10. b. for the answer.

In similarway, we can getthe probability of the event that the last ball in the list will be red is

\(\begin{aligned}{c}{P_{99 + 1}} &= {P_{100}}\\ &= \frac{k}{n}\\ &= \frac{r}{{100}}\end{aligned}\).