Solve Probability Problem: An Urn with 5 Black & 4 White Balls

[Yankel]

An urn has 5 black balls ans 4 white balls in it. We randomly choose a ball, and return it to the urn, until we get 2 balls with the same colour. What is the probability that the first ball was white, if we know that the last one was white ? I tried building a tree, and realized that the experiment can have 2 or 3 stages, not more. The answer say 65/81, which makes no sense to me, I think it's a mistake. Can you please help me solve this problem ? I used conditional probability, with no luck.

Thanks !

 

[I like Serena]

An urn has 5 black balls ans 4 white balls in it. We randomly choose a ball, and return it to the urn, until we get 2 balls with the same colour. What is the probability that the first ball was white, if we know that the last one was white ? I tried building a tree, and realized that the experiment can have 2 or 3 stages, not more. The answer say 65/81, which makes no sense to me, I think it's a mistake. Can you please help me solve this problem ? I used conditional probability, with no luck.

Thanks !

Hey Yankel! ;)

Possible outcomes that end with a white ball, are:
\begin{array}{|c|c|}
\hline
\text{Outcome} & \text{Unconditional Probability} \\
\hline
WW & \frac 4{11} \cdot \frac 3{10} \\
B\,WW & \frac 5{11} \cdot \frac4{10} \cdot \frac 3{9} \\
WB\,WW & ...\\
B\,WB\,WW \\
WB\,WB\,WW \\
B\,WB\,WB\,WW \\
\hline
\end{array}
And we're looking for:
$$P(\text{first white} \mid \text{last white}) = \frac{P(\text{first white} \wedge\text{last white})}{P(\text{last white})}$$
(Thinking)

 

[Yankel]

I did not understand the experiment to end when there are two balls of the same color in a row...
This is an interesting interpretation, maybe this is my mistake. I though you could not get more than 2 W overall.

And why dividing by 11 ? There are 9 balls.

 

[I like Serena]

I did not understand the experiment to end when there are two balls of the same color in a row...
This is an interesting interpretation, maybe this is my mistake. I though you could not get more than 2 W overall.

And why dividing by 11 ? There are 9 balls.

Hmm... your interpretation could well be right as well... (Thinking)
In that case, we would only have WW, WBW, and BWW.
Still, if I calculate that, I'm getting $P(\text{first white} \mid \text{last white}) = \frac {12}{17}$.

And you are quite right, the division should be by 9 instead of by 11.