Q:

Probability of grabbing blue ball

We have a bag that has

$3$ blue balls and $2$ white balls.

When we take a blue ball from the bag, we just remove it from the bag.
When we take a white ball from bag, we return that ball and add another white ball.
What is the probability that on our first take we take a blue ball when we know that after $2$ takes the amount of balls in the bag has changed?
My approach:
We know that amount of balls in bag changed after $2$ takes , so:
$( B , B ) = -2$ balls
$( B , W ) = 0$ balls
$( W, W ) = 2$ balls
$( W, B ) = 0$ balls
the only acceptable order is $BB$ or $WW$.
And now I am confused about how to decide what is the probability that on our first take we took a blue ball?
The probability for blue ball is $3/4$ and when we take that ball we have $2/4$ chance of another blue ball.
So is it $(3/4) \cdot ( 1/2)$
OR should I take in considering that we have $1/2$ chance, that $( B, B )$ occurs, and then multiply it by $3/4$ ?
I know that in this case, both results are the same, but if we had a different amount of balls, it would differ, and I am not sure which approach is correct.
Thanks for your help.

A:

Let $E$ denote the event that after two takes the number of balls in the bag
has changed.
Let $B_{i}$ denote the event that at the $i$-th take a blue ball
is taken, and let $W_{i}$ denote the event that at the $i$-th take
a white ball is taken.
Then to be found is $P\left(B_{1}\mid E\right)$
We have the relation:
$$P\left(B_{1}\mid E\right)P\left(E\right)=P\left(B_{1}\cap E\right)\tag1$$
So after finding $P\left(E\right)$ and $P\left(B_{1}\cap E\right)$
we can also find $P\left(B_{1}\mid E\right)$.
First we will go for finding $P(E)$:
$$P\left(E\right)=P\left(\left(B_{1}\cap B_{2}\right)\cup\left(W_{1}\cap W_{2}\right)\right)=P\left(B_{1}\cap B_{2}\right)+P\left(W_{1}\cap W_{2}\right)=$$$$P\left(B_{2}\mid B_{1}\right)P\left(B_{1}\right)+P\left(W_{2}\mid W_{1}\right)P\left(W_{1}\right)=\frac{2}{4}\frac{3}{5}+\frac{3}{6}\frac{2}{5}=\frac{1}{2}$$
Now we will go for finding $P\left(B_{1}\cap E\right)$:
$$P\left(B_{1}\cap E\right)=P\left(B_{1}\cap B_{2}\right)=P\left(B_{2}\mid B_{1}\right)P\left(B_{1}\right)=\frac{2}{4}\frac{3}{5}=\frac{3}{10}$$
Substituting these results in $(1)$ we find: $$P\left(B_{1}\mid E\right)\frac{1}{2}=\frac{3}{10}$$
and we conclude that: $$P\left(B_{1}\mid E\right)=\frac{3}{5}$$

