91Ó°ÊÓ

Initially,

Number of red balls = r

Number of blue balls = b

Each time a ball is selected on any random trial, the ball along with k additional balls of the same color are put into the box.

Let:

\({R_1} = \)Event that the first selected ball is red

\({R_2} = \)Event that the second selected ball is red

\({R_3} = \)Event that the third selected ball is red

\({B_4} = \) Event that the fourth selected ball is blue

Initially, when the box contained rred balls and b blue balls, the probability that the first ball is red is:

\(P\left( {{R_1}} \right) = \frac{r}{{r + b}}\)

The first selected red ball is returned to the box and k additional red balls are put into the box after the first draw.

Using the Multiplication Rule for Conditional Probabilities,

\({\bf{Pr}}\left( {{\bf{A}} \cap {\bf{B}}} \right){\bf{ = Pr}}\left( {\bf{A}} \right){\bf{ Pr}}\left( {\left. {\bf{B}} \right|{\bf{A}}} \right)\)

Thus,

\(\begin{aligned}{}P\left( {{R_1} \cap {R_2}} \right) &= P\left( {{R_1}} \right) \times P\left( {{R_2}|{R_1}} \right)\\ &= \frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}}\end{aligned}\)

Similarly, the second selected red ball is returned to the box and k additional red balls are put into the box after the second draw.

Thus,

\(\begin{aligned}{}P\left( {{R_1} \cap {R_2} \cap {R_3}} \right) &= P\left( {{R_1} \cap {R_2}} \right) \times P\left( {{R_3}\left| {{R_1} \cap {R_2}} \right.} \right)\\ &= \frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}} \times \frac{{\left( {r + k} \right) + k}}{{\left( {r + k + b} \right) + k}}\\ &= \frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}} \times \frac{{r + 2k}}{{r + 2k + b}}\end{aligned}\)

Again, the second selected red ball is returned to the box and k additional red balls are put into the box after the third draw.

Finally, the probability that the fourth ball will be blue is obtained as:

\(\begin{aligned}{}P\left( {{R_1} \cap {R_2} \cap {R_3} \cap {B_4}} \right) &= P\left( {{R_1} \cap {R_2} \cap {R_3}} \right) \times P\left( {{B_4}|{R_1} \cap {R_2} \cap {R_3}} \right)\\ &= \frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}} \times \frac{{r + 2k}}{{r + 2k + b}} \times \frac{b}{{\left( {r + 2k + b} \right) + k}}\\ &= \frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}} \times \frac{{r + 2k}}{{r + 2k + b}} \times \frac{b}{{r + 3k + b}}\end{aligned}\)

Therefore, the probability that the first three balls will be red and the fourth ball will be blue is \(\frac{r}{{r + b}} \times \frac{{r + k}}{{r + k + b}} \times \frac{{r + 2k}}{{r + 2k + b}} \times \frac{b}{{r + 3k + b}}\)