the probability that it is a boy knowen that is not from second class 


A survey conducted in the Documentation and Information Centre (CDI)
  of a high school gives the following results: 60% of students
  frequenting the CDI are girls and among them, 40% are in the second,
  30% in the first and the remaining terminal. Among boys frequenting
  the CDI, 50% are in the second, 20% in the first one and the rest in
  the terminal.
Were randomly questioned a high frequenting the CDI and we consider
  the following events:
F: "the pupil is a girl asks," G: "asks the pupil is a boy" S: "the
  pupil is questioned in second" P: "is the first high interrogates" T:
  "asks the pupil is terminal."
Question 5- The pupil asks not to be from the second class . so what
  is  the probability $P_{4}$ whether a boy

beware if you don't understand my translation from the source you should wait until someone will translate it  i tag translation-request


*

*I want to understand how they did to get 


$$P_{4}=0.3571$$
my attempts
since the student is not in second class so it is first or terminal class ....
Let E be the set of first or  terminal class
$P(E)=P(F\cap  P )P(F\cap  T )+P(G\cap  P )P(G\cap  T )$ 
the probability that it is a boy not in the second class
$$P_{E}(G) = \dfrac {P(E \cap G)}{P(E)}$$
 A: Nevertheless, here is the translation, not the answer...
In the Information Center (IC) of a high school a survey is conducted and gives the following results:


*

*$60$% of students in the IC are girls among which $40$% are "secondes" (freshmen), $30$% are "premières" (sophomore) and the rest "terminales" (senior) (only three levels in French high schools)

*Among boys in the IC, $50$% are secondes, $20$% premières and the rest terminales
We are taking one student at random from student in the IC and here are the possible events:
F: the student is a girl; G: the student is a boy; S: the student is a seconde; P: the student is a première; T: the student is a terminal
I-A-1 Complete the given tree with corresponding probabilities
I-A-2 Give probability $P_1$ that the student is a girl in seconde
I-A-3 Give probability $P_2$ that the student is a seconde
I-A-4 The student is a seconde. Give the probability $P_3$ that this is a girl. Explain your answer. Give an approximation up to $10^{-4}$.
I-A-5 The student is not a seconde. Give the an approximation up to $10^{-4}$ of the probability $P_5$ that the student is a boy.
A: This is an example of conditional probability.  The question asks us to find $Pr(G|S^c)$, i.e. the probability that the person being questioned is in fact a boy (garçon) given that they were not from the second class.
By definition of conditional probability:
$$Pr(G|S^c) = \frac{Pr(G\cap S^c)}{Pr(S^c)}$$
The probability $Pr(G\cap S^c)$ can be calculated from the above tree diagram as the summation of all probabilities of leaves corresponding to the event.  I.e. $$Pr(G\cap S^c) = Pr(G\cap P) + Pr(G\cap T) = Pr(G)\cdot Pr(P|G) + Pr(G)\cdot Pr(T|G) \\= 0.4\cdot 0.2 + 0.4\cdot 0.3 = .08 + .12 = .20$$
The probability $Pr(S^c)$ can be calculated similarly.  Add up the probabilities of each leaf corresponding to $S^c$.
$$Pr(S^c) = Pr(F\cap P)+ Pr(F\cap T) + Pr(G\cap P) + Pr(G\cap T)\\
= .6\cdot .3 + .6\cdot .3 + .4\cdot .2 + .4\cdot .3 = .36+.20 = .56$$
For a final probability of:
$$\frac{Pr(G\cap S^c)}{Pr(S^c)} = \frac{.20}{.56} =\frac{5}{14}\approxeq 0.3571$$

minor notes: my notation $Pr(A|B)$ seems to be equivalent to your notation $P_B(A)$, and my notation $A^c$ refers to the complement event, $A^c = \Omega\setminus A$, i.e. my $S^c$ corresponds to your $E$ your shown attempt.
