Probability of passing a test: theory test and practice test. Dependent or independent events? context
I was doing a math exam simulation and I found a problem on probability that I can't understand.
I checked the self-assessment grid and then went back to my textbook avoiding wikipedia or more advanced material on purpose in order to understand exactly what they expect me to answer with the knowledge they assume I have (math exam IV° EU level if can be of any help). Sadly I can't ask my teacher.

The question is${}^{*}$ the following

There are two parts in a driving test: a theory test and a
  practice test.
To do the practice test you have to pass the theory test.
The probability to pass the theory test is $0.8$; the probability to pass the practice test is $0.6$.
$\mathcal Q\quad$ What is the probability of successfully pass the driving test?

($*$ [bad] translation, emphasis and bold are mine)

$\mathcal A\quad$ The self-assessment grid tells me that the answer is $0,48$


At first I tried to no overthink it and went like this: let 
$T:=\text{"I pass the theory test"}$
$P:=\text{"I pass the practice test"}$
${\bf D}:=\text{"I pass the drive test"}$
By the definition given in the first paragraph we have that passing the drive test is equivalent to $T$ and $P$ happening: ${\bf D}$ happens if and only if $T$ and $P$ happens ). Maybe with an abuse of notation I'd write this as follows:
$${\bf D}\iff T\land P$$
Since passing $T$ or passing $P$ are independent (?), and denoting with $p(E)$ the probability of an event $E$ we "maybe" have
$$p({\bf D})=p(T\land P)=p(T)\cdot p(P)$$
So the result is $0.8 \cdot 0.6=0.48$
Well... this should be correct but during the simulation it didn't sounds correct to me so I went ahead and gave another (wrong?) answer.

What made me think was the second paragraph of the question... it actually tells that If I even want to try the practice test I have to pass the theory test first... so the two events are not exactly independent... does this change something?
My argument go as follows: paragraph 2 says that "if $P$ happens then $T$ happened" so it says $$P\implies T$$
But if this is right then "if $P$ happens the also $T$ happened" and, by definition, $\bf D$ happens, i.e. "I pass the practice test" is enough for "passing the whole driving test". 
$$P\implies \bf D$$
But since ${\bf D}=T\land P$ implies by definition that both $T$ and $P$ happened (are true) then $$ {\bf D}\implies {P}$$...
I'm tempted to conclude that $p({\bf D})=p(P)=0.6$ because tha probability of $P$ should already contain the probability of passing the theory.
$...$
There is still something that seems missing but I also have to admit that "probably" there are some gaps in my understanding of probability theory test.


$\mathcal Q$uestions: 
  
  
*
  
*My first argument seems imprecise but it gives exactly the answer they wanted... is it correct?
  
*What is wrong in my second argument and what really is the correct solution?


I apologize for my bad grammar and bad English... if my errors hurt your eyes to much you can edit tem or post a comment and I'll edit asap
 A: You're right.  Technically, for the question to be accurate it should ask "the probability to pass the practice test, given that you take it, is 0.6".  Since of the $0.4$ that do not pass it, in reality, without that proviso, some of those who do not pass it will be those who do not take it because they failed the theory.
Furthermore, they assume the two are independent which should also be stated in the interests of an accurate question.  In the real world, there is almost certainly a positive correlation between competence in one field and competence in the other.
If you are meticulous in your attention to detail, which is a good quality in maths, you will frequently find situations like this where your understanding of a question goes beyond what a careless question-writer intended and you have to make allowances for them.
A: There are no independent events present. We have two probabilities $\mathbb P(T)=0.8$ and $\mathbb P(\overline T)=0.2$ where $\overline T$ is the event opposite to $T$. And also we have two conditional probabilities $\mathbb P(P\mid T)=0.6$ and $\mathbb P(P\mid \overline T)=0$. This zero value is due to the words 

To do the practice test you have to pass the theory test.

If $T$ and $P$ are independent then $\mathbb P(P\mid T)=\mathbb P(P\mid \overline T)$. But the l.h.s. equals to $0.6$ while r.h.s. equals to zero. So, these events are dependent. Moreover, $P\subseteq T$ since if $P$ occures then $T$ occures too. 
The event $D$ is intersection of events $T$ and $P$. That is, it occures when both this events occure:  $D=T\cap P$.
By definition of conditional probability, 
$$\mathbb P(P\mid T)=\dfrac{\mathbb P(T\cap P)}{\mathbb P(T)}$$
which implies 
$$\mathbb P(D)=\mathbb P(T\cap P)=\mathbb P(T)\cdot\mathbb P(P\mid T)=0.8\cdot 0.6.$$
The probability of both events $T$ and $P$ occured is the product of probability of the first one and the conditional probability of the second one if the first occured. 
The unconditional probability that the practice test will be passed by some individual 
$$\mathsf P(P)=\mathbb P(D)=0.48$$ 
since $P\subseteq T$, $P\cap T=D$.
In order to understand the relations of this events and probabilities correctly, It is convenient to imagine a lot of people ($100\%$) going to take exams for a driver's license. $T$ is a part of those people who will pass theoretical exam ($80\%$). Among these people $60\%$ will pass a practical exam. You can see that $60\%$ refers only to those people who pass the theoretical exam. And $0\%$ will pass practice among those people who did not pass theoretical exam. The total proportion of people who pass practical exam will be $0.8\cdot 0.6=0.48$ or $48\%$. This is $60\%$ of  $80\%$. 
