#33 is 0.3 times 42, because 0.3 is basically 30%, and you always multiply the percent by the number to get the answer. It could also be 3/10 times 42.

3 0

3 years ago

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 ex

s344n2d4d5 [400]

Answer:

Suppose the fraction of blue balls at any given state n is $f_{n}$ . At the start of the process we have: $f_{0}$ =1.

Now consider the situation after having drawn n times.

The current fraction of blue balls is $f_{n}$ , the number of blue balls is therefore 10 $f_{n}$ .

We either draw a red ball, with probability 1− $f_{n}$ . and the number of red balls, blue balls stay the same (we swap red for red):

$f_{n+1}$ (red)= $f_{n}$

or we draw a blue ball, with probability $f_{n}$ and we obtain 10 $f_{n}$ −1 blue balls on a total of 10 balls, thus:

$f_{n+1}$ (blue) = (10 $f_{n}$ − 1)/10

The total probability of (fraction of) blue balls after this n'th draw is therefore:

$f_{n+1}$ = $f_{n+1}$ (red)(1− $f_{n}$ )+ $f_{n+1}$ (blue) $f_{n}$

which can be simplified to:

$f_{n+1}$ =0.9 $f_{n}$

which leads to:

fn=0.9 $f_{n}$

At the end of the 10'th draw the fraction of blue balls is equal to:

$f_{10}$ = $0.9^{10}$ ≈ 0.348678

8 0

2 years ago