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Imagine a bag filled with 12 red balls 12 white balls 12 blue balls

Every time one ball is drawn from the bag and after each drawn, the ball would be kept back in the bag so ball numbers doesn't change.

The question is: if we don't see any white ball drawn in first 10 draw, what is the possibility of a white ball at 11th drawn? Does it still 1/3 or gets higher by the drawn count?

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It doesn't change. Since balls are replaced, the draws are clearly independent and so consider that everytime we draw a ball, the probability of getting a white ball is $p$ and the probability of not getting one is $1-p$.

Now, if $X$ is the number of attempts until we get a white ball then $X$ follows a geometric distribution. The geometric distribution is the discrete version of the exponential distribution, and so it also possesses the memoryless property, the idea that past attempts do not affect future attempts.

Let $S$ be the event we have unsuccessfully tried to draw a white ball $s$ times, and $Y$ the event we draw a white ball on the next attempt. Then:

$$P(Y | S) = \frac{P(Y \cap S)}{P(S)}$$ $$= \frac{P(X=s+1)}{P(S)}$$ $$=\frac{p (1-p)^s}{(1-p)^s}$$ $$=p$$

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    $\begingroup$ Thanks for the answer $\endgroup$ – Rasko Jul 28 '17 at 23:39

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