I don't know where I went wrong.
Here the part of the math problem:
We have two boxes, U1 and U2
U1 has:
(n) number of white balls.
(3) number of black balls
U2 has:
(2) number of white balls.
(1) number of black balls
We do experiment E:
We take, randomly, 1 ball from U1 and we put it in U2. Then, we take 1 ball from U2 and we put it in U1
1) We consider the A probability: "After the experiment E, the two boxes stay like the way they were"
A)Prove that the probability p(A) is written like this: p(A) = (3/4) * ( (n+2) / (n+3) )
And here is my solution:
Since the two boxes stay the way they are. That means we have to take from each box 1 ball of the same color as the other so that the number of black balls and white balls stay the same.
So we calculate the number of probabilities for each box:
The first box: $P(1,n+1) = n+3$
The second box: P(1,4) = 12 => I thought that after putting a ball in U2 from U1, they will be four balls.
Then I calculated like this:
P(A) = ([P(1,3)/n+3] * [P(1,2)/12]) + ([P(1,n)/n+3] * [P(1,3)/12)]
But I found it to be P(A) = (3/4) * ( (n+2) / 3(n+3) )
Why 3 is in there? What I have done wrong?
Btw, I used [] so that I make it much easier to read and give priorities to the operations.