Difficulties in understanding a conditional probability question from a textbook In the book "Complete Probability & Statistics 1 for Cambridge International AS & A Level" I found the following question:

The homework diaries and completed homework of two Students A and B,
are examined.
There is a probability of 0.4 that Student A does not make a note in
her diary homework set. She always does the homework if it is written
in her diary, but never does the homework if it is not written in.
There is a probability of 0.8 that Student B writes the homework given
in her diary. When she does this, she will do the homework 90% of the
time. If she has nothing written in her diary then she checks with a
friend, who knows what the homework is 50% of the time; Student B
always does the homework if she is told what it is by her friend.
A) Find the probability that Stundent B does her homework on a particular
night.
B) Find the probability that both students do her homework on a
particular night.
C) If one piece of homework was given to a student
but wasn’t done, find the probability that it was given to Student A.

For A) I got $0.8\cdot 0.9+0.2\cdot 0.5=0.82$.
For B) I got $0.82\cdot 0.6=0.492.$
For C) I got: The probability that the homework is not done by both or one student is $1-0.492=0.508$. Therefore we find that $\frac{0.4}{0.508}\approx 0.7874$
But consulting this forum I found that the solution is apparently $\frac{0.382}{0.382+0.108}=0.752$
I do not understand why this is correct. For me, 0.752 is the solution to the question:  If homework of both was checked and one was not done, find the probability that it was Student A's.
Can somebody tell me if I am missing something?
 A: 

The homework diaries and completed homework of two Students A and B,
are examined.
There is a probability of 0.4 that Student A does not make a note in
her diary homework set. She always does the homework if it is written
in her diary, but never does the homework if it is not written in.
There is a probability of 0.8 that Student B writes the homework given
in her diary. When she does this, she will do the homework 90% of the
time. If she has nothing written in her diary then she checks with a
friend, who knows what the homework is 50% of the time; Student B
always does the homework if she is told what it is by her friend.
C) If one piece of homework was given to a student but wasn’t done,
find the probability that it was given to Student A.

I got: The probability that the homework is not done by both or one
student is $1-0.492=0.508$. Therefore we find that
$\frac{0.4}{0.508}\approx 0.7874.$

Firstly, you are interpreting "one student" as "at least one student", whereas the author probably means "exactly one student".
Secondly, you aren't taking into account the given condition that one piece of homework wasn't done.
Now, P(A did but not B) $= [0.6 \times (0.8\times0.1+0.2\times0.5)]=0.108,$
while P(B did but not A) $= [(1-0.6) \times (1-(0.8\times0.1+0.2\times0.5))]=0.328.$
The required probability is P(B did but not A) $\div$  [P(A did but not B) + P(B did but not A)], which equals $$\frac{0.328}{0.328+0.108}=0.752.$$

But consulting this forum I found that the solution is apparently
$\frac{0.382}{0.382+0.108}=0.752.$
I do not understand why this is correct. For me, 0.752 is the solution
to the question:  If homework of both was checked and one was not
done, find the probability that it was Student A's.

Exactly! But, to be exact, "exactly one was not done". -)
By the way, your transcribed answer from that forum contains two identical typos.
