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

Question

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

$$P_{E}(G) = \dfrac {P(E \cap G)}{P(E)}$$

Answer 1

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.

Answer 2

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:

$$\frac{Pr(G\cap S^c)}{Pr(S^c)} = \frac{.20}{.56} =\frac{5}{14}\approxeq 0.3571$$

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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.