Let $n$ be a natural number, $p$ a real number between 0 and 1, and $X_n$ a random variable following a binomial distribution with parameters $n$ and $p$. We denote $F_n = \frac{X_n}{n}$ and $f$ a value taken by $F_n$. We recall that, for $n$ sufficiently large, the interval $\left[p - \frac{1}{\sqrt{n}} ; p + \frac{1}{\sqrt{n}}\right]$ contains the frequency $f$ with probability at least equal to 0.95.
Deduce that the interval $\left[f - \frac{1}{\sqrt{n}} ; f + \frac{1}{\sqrt{n}}\right]$ contains $p$ with probability at least equal to 0.95.

We seek to study the number of students knowing the meaning of the acronym URSSAF. For this, we survey them by proposing a multiple choice questionnaire. Each student must choose from three possible answers, denoted $A$, $B$ and $C$, the correct answer being $A$. We denote by $r$ the probability that a student knows the correct answer. Any student knowing the correct answer responds $A$, otherwise they respond at random (equiprobably).

1. We survey a student at random. We denote: $A$ the event ``the student responds $A$'', $B$ the event ``the student responds $B$'', $C$ the event ``the student responds $C$'', $R$ the event ``the student knows the answer'', $\bar{R}$ the complementary event of $R$. a. Translate this situation using a probability tree. b. Show that the probability of event $A$ is $P(A) = \frac{1}{3}(1 + 2r)$. c. Express as a function of $r$ the probability that a person who chose $A$ knows the correct answer.
2. To estimate $r$, we survey 400 people and denote by $X$ the random variable counting the number of correct answers. We will assume that surveying 400 students at random is equivalent to performing sampling with replacement of 400 students from the set of all students. a. Give the distribution of $X$ and its parameters $n$ and $p$ as a function of $r$. b. In an initial survey, we observe that 240 students respond $A$, out of 400 surveyed. Give a 95\% confidence interval for the estimate of $p$. Deduce a 95\% confidence interval for $r$. c. In what follows, we assume that $r = 0.4$. Given the large number of students, we will consider that $X$ follows a normal distribution. i. Give the parameters of this normal distribution. ii. Give an approximate value of $P(X \leqslant 250)$ to $10^{-2}$ precision.

The store manager wishes to estimate the proportion of defective padlocks in his stock of budget padlocks. For this, he takes a random sample of 500 budget padlocks, among which 39 prove to be defective.
Give a confidence interval for this proportion at the $95\%$ confidence level.

We denote by $p$ the unknown proportion of young people aged 16 to 24 years who practice illegal downloading on the internet at least once a week.
A young person participating in protocol $( \mathscr { P } )$ is randomly selected. The protocol $( \mathscr { P } )$ is as follows: each young person rolls a fair 6-sided die; if the result is even, the young person answers sincerely; if the result is ``1'', the young person must answer ``Yes''; if the result is ``3 or 5'', the young person must answer ``No''.
We denote: $R$ the event ``the result of the roll is even'', $O$ the event ``the young person answered Yes''.

1. Probability calculations

Reproduce and complete the weighted tree diagram.
Deduce that the probability $q$ of the event ``the young person answered Yes'' is: $$q = \frac { 1 } { 2 } p + \frac { 1 } { 6 }$$

2. Confidence interval

a. At the request of Hadopi, a polling institute conducts a survey according to protocol $( \mathscr { P } )$. On a sample of size 1500, it counts 625 ``Yes'' responses. Give a confidence interval, at the 95\% confidence level, for the proportion $q$ of young people who answer ``Yes'' to such a survey, among the population of young French people aged 16 to 24 years. b. What can be concluded about the proportion $p$ of young people who practice illegal downloading on the internet at least once a week?