A conditional probability question The question is:
There are two box, the first one contains n red balls, and the second one contains m balls with only one red ball and otherwise blue, (m, n>2). Now we randomly select one box, then choose two balls in turn without replacement.
(1). Find out the probability p with which the ball chosen first is red
(2). Under the condition in which the first ball is red, calculate the probability q with which the second ball is also red.
Evidently, 
                               p = (m+1)/2m

However, the probability q which I calculated is 
                               p = m/(m+1) 

is wrong as the author said it's a common error. Correct answer the author gave is 
                                q = 1/2

I read the comment given by the author. Let Ai(i=1,2) denotes the event of (i)th box being selected, C denotes the event the ball chosen is red. The author said that 
                             P(A1) = 1/2
                             P(A2) = 1/2
                             P(C|A1) = 1
                             P(C|A2) = 0

Thus,
                       q = P(C) = 1*1/2 + 0*1/2

And the wrong algorithm I used (Bi denotes the event that (i)th ball chosen is red)
                             q = P(B2|B1)

The author pointed out that it cannot guarantee that the partitioned B1 be the same in the numerator and denominator if we continue this method.
However, I thought this question in another way. If the first ball chosen is red, the probability of the selected box is first one or second one will change as I think.
Therefor, I calculated the probability
                              P(A1|B1) = m/(m+1)
                              P(A2|B1) = 1/(m+1)

Thus, I think that the probability of the second ball chosen being red is 
                           q=1*m/(m+1) + 0*1/m+1

which is the same as my precedent wrong answer.
My question is what's the fallacy of my reasoning. And could you show me  a restricted mathematical calculation of the answer not just verbal reasoning?
Because question is hard to google, I'm sorry for duplicating. If there are a similar question. Please give me a pointer. Thank you! 
English is not my native language, if I said something which is inaccurate . Please point me out. Any help will be appreciated. Thanks!
 A: Let $A_i$ be the event that the $i$-th box is chosen, $B$ and $C$ be the event that the $1$st and $2$nd ball is red respectively. 
First, a side issue here is that the author has oppressed the conditional notation, so the probability $P$ used in the comment is different from the 1st part which will easily cause confusion. What he mean is
$$ P(C|A_1, B) = 1, P(C|A_2, B) = 0$$
Note that without the information that the $1$st ball is red, we have
$$ P(C|A_2) = P(B|A_2) = \frac {1} {m}$$
Now the main issue is how do you interpret $q$ - what is the meaning of "Under the condition in which the first ball is red"
The author mean that
$$ q = P(B \cap C) = P(B \cap C|A_1)P(A_1) + P(B \cap C|A_2)P(A_2) = \frac {1} {2} $$
But you interpret as
$$ q = P(C|B) = \frac {P(B \cap C)} {P(C)} = \frac {1} {2} \times \frac {2m} {m+1} = \frac {m} {m+1}$$
So this cause the difference. The lexical ambiguity here is not a big issue in your study - as long as you know the math it is ok. IMHO I will vote for your answer.
A: $P(R,R) = P(B1).P(2nd Red/1st Red) + P(B2).P(2nd Red/1st Red) = \frac{1}{2}.1 + \frac{1}{2}.0 = \frac{1}{2}$
 Your answer is incorrect because, the first selection  is again amongst the boxes and then two balls, thus, the first box contains all red and hence P(2nd is Red/1st is Red in Box1) = 1 and in the other box P(2nd is Red /1st is Red in Box 2) = 0(due to without replacement condition)
Did you not use the above logic for the first part.
$P(R) = P(box 1) . P(red/Box 1) + P(box 2).P(red/Box 2) = \frac{1}{2}.1 + \frac{1}{2}.\frac{1}{m}$.
