The solution guide says that P(X=n) = 1/2 * 2/3 * 3/4 …
But I don't understand how P(X=2) = 2/3 if we have 1 black ball and 2 white balls at the second stage? Shouldn’t that be 1/3?
A box has 2 balls, one B and one W. We draw one randomly. If it's B, we stop. If it's W, we put it back in with another W ball. What's P(X=n)?
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probability
2 Answers
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$\mathsf P(X=2)$ is the probability for drawing one white ball, replacing it and adding an extra, then drawing the black ball. $\tfrac 12\tfrac 13 =\tfrac 16$. And likewise for any $k$ :-
So, you want the probability for drawing the black ball for the first time on draw number $k$. That is, for drawing $k-1$ consecutive white balls, while incrementing the counts of white (and total) balls by one each draw, then finally drawing the lone black ball from the $k+1$ balls in box at that time.
$$\begin{align}\mathsf P(X=k) ~&=~\frac{1}{2}\cdot\frac{2}{3}\cdot \frac 34\cdots\frac{k-1}{k}\cdot\frac 1{k+1} \\[1ex] & = \frac{1}{k\,(k+1)}\\[3ex] \mathsf P(X=1) & = \frac 12 \\[1ex] \mathsf P(X=2) & = \frac 1 6\\[1ex]\mathsf P(X=3) & = \frac 1 {12}\end{align}$$
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P(X=n)=P1 (W)P2(W)*P3(W)...Pn(B)
Since to draw B ball by the n time it is required to not have drawn the B ball all the previous n-1 times