Answer:
Suppose the fraction of blue balls at any given state n is
. At the start of the process we have:
=1.
Now consider the situation after having drawn n times.
The current fraction of blue balls is
, the number of blue balls is therefore 10
.
We either draw a red ball, with probability 1−
. and the number of red balls, blue balls stay the same (we swap red for red):
(red)=
or we draw a blue ball, with probability
and we obtain 10
−1 blue balls on a total of 10 balls, thus:
(blue) = (10
− 1)/10
The total probability of (fraction of) blue balls after this n'th draw is therefore:
=
(red)(1−
)+
(blue)
which can be simplified to:
=0.9
which leads to:
fn=0.9
At the end of the 10'th draw the fraction of blue balls is equal to:
=
≈ 0.348678