<s>[INST] What is the probability of drawing the last red ball in the fourth draw? There are 8 Blue(B) and 2 Red(R) balls in a bag. Each time one ball is drawn and replaced by Blue one. What is the probability of drawing the last red ball in the fourth draw ? 

I made 3 cases :
A). BBRR = $(\frac{8}{10})^2\cdot \frac{2}{10}\cdot \frac{1}{10}$
B). RBBR = $\frac{2}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{1}{10}$
C). BRBR = $\frac{8}{10}\cdot \frac{2}{10}\cdot \frac{9}{10}.\frac{1}{10}$
Hence, Total Probability :-
=> $P(A) + P(B) + P(C)$
=> ${ (\frac{8}{10})^2\cdot \frac{2}{10}\cdot \frac{1}{10} }$ + $\frac{2}{10}\cdot \frac{9}{10}\cdot \frac{9}{10}\cdot \frac{1}{10}$ + $\frac{8}{10}\cdot \frac{2}{10}\cdot \frac{1}{10}$

Have I made it Right ?
 [/INST] Just to generalize a little bit 
Assuming $ N $ = Total Number of draws 
Notes: 


*

*Probability to draw the first red ball is always 2/10 

*Probability to draw the second red ball is always 1/10 


If $ i \in (1,N-1) $ is the draw you get the first red ball at, then 


*

*Probability to draw a blue ball for $ 1 \le j < i $ is 8/10 

*Probability to draw a blue ball for $ i < j < N-1 $ is 9/10 


Then you can define a function for computing the probability of a specific scenario 
$$ P(i;N) = \left(\frac{8}{10}\right)^{i-1} \left(\frac{9}{10}\right)^{N-1-i} \left(\frac{2}{10}\right) \left(\frac{1}{10}\right) $$
Then you can compute the sum over all the possible scenario 
$$ P_{final} = \sum_{i=1}^{N-1}P(i;N) $$
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