Starting with 10 blue balls, in each of 10 sequential rounds, we remove a random ball and replace it with a new red ball. For example, after the first round we have 9 blue balls and one red ball, after the second round, with probability 9/10 we have 8 blue balls and 2 red balls, and with probability 1/10 we have 9 blue balls and one red ball, etc. What is the probability that the ball we remove at the 11th round is blue?

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

14 x 10 5

step-by-step explanation:

10 5 = 100,000

14x10,000 = 140,000