To conduct a survey, an employee interviews people chosen at random in a shopping mall. He wonders whether at least three people will agree to answer.

1. In this question, we assume that the probability that a person chosen at random agrees to answer is 0.1. The employee interviews 50 people independently. We consider the events: $A$: ``at least one person agrees to answer'' $B$: ``fewer than three people agree to answer'' $C$: ``three or more people agree to answer''. Calculate the probabilities of events $A$, $B$ and $C$. Round to the nearest thousandth.
2. Let $n$ be a natural integer greater than or equal to 3. In this question, we assume that the random variable $X$ which, to any group of $n$ people interviewed independently, associates the number of people who agreed to answer, follows the probability distribution defined by: $$\left\{\begin{array}{l}\text{For every integer } k \text{ such that } 0 \leqslant k \leqslant n-1,\; P(X = k) = \frac{\mathrm{e}^{-a} a^k}{k!}\\\text{and } P(X = n) = 1 - \sum_{k=0}^{n-1} P(X=k)\end{array}\right.$$

A factory of frozen desserts has an automated line to fill ice cream cones. Ice cream cones are packaged individually and then packaged in batches of 2000 for wholesale sale. It is considered that the probability that a cone has any defect before its packaging in bulk is equal to 0.003. We denote by $X$ the random variable which, to each batch of 2000 cones randomly selected from production, associates the number of defective cones present in this batch. It is assumed that the production is large enough that the draws can be assumed to be independent of each other.

1. What is the distribution followed by $X$? Justify the answer and specify the parameters of this distribution.
2. If a customer receives a batch containing at least 12 defective cones, the company then proceeds to exchange it. Determine the probability that a batch is not exchanged; the result will be rounded to the nearest thousandth.

Question 2

In this hypermarket, a computer model is on promotion. A statistical study made it possible to establish that, each time a customer is interested in this model, the probability that they buy it is equal to 0.3. We consider a random sample of ten customers who were interested in this model. The probability that exactly three of them bought a computer of this model has a value rounded to the nearest thousandth of: a. 0.900 b. 0.092 c. 0.002 d. 0.267

Part A

A competitor participates in an archery competition on a circular target. With each shot, the probability that he hits the target is equal to 0.8.

1. The competitor shoots four arrows. It is considered that the shots are independent. Determine the probability that he hits the target at least three times.
2. How many arrows should the competitor plan to shoot in order to hit the target an average of twelve times?

On a tennis court, a ball launcher allows a player to train alone. This device sends balls one by one at a regular rate. The player then hits the ball and the next ball arrives. According to the manufacturer's manual, the ball launcher sends the ball randomly to the right or to the left with equal probability.
Throughout the exercise, results will be rounded to $10 ^ { - 3 }$ near.

Part A

The player is about to receive a series of 20 balls.

1. What is the probability that the ball launcher sends 10 balls to the right?
2. What is the probability that the ball launcher sends between 5 and 10 balls to the right?

Part B

The ball launcher is equipped with a reservoir that can hold 100 balls. Over a sequence of 100 launches, 42 balls were launched to the right. The player then doubts the proper functioning of the device. Are his doubts justified?

Part C

To increase the difficulty, the player configures the ball launcher to give spin to the balls launched. They can be either ``topspin'' or ``slice''. The probability that the ball launcher sends a ball to the right is still equal to the probability that the ball launcher sends a ball to the left. The device settings allow us to state that:

• the probability that the ball launcher sends a topspin ball to the right is 0.24;
• the probability that the ball launcher sends a slice ball to the left is 0.235.

If the ball launcher sends a slice ball, what is the probability that it is sent to the right?

The different sweets in the bags are all coated with a layer of edible wax. This process, which deforms some sweets, is carried out by two machines A and B. When produced by machine A, the probability that a randomly selected sweet is deformed is equal to 0.05.
On a random sample of 50 sweets from machine A, what is the probability, rounded to the nearest hundredth, that at least 2 sweets are deformed?
Answer a: 0.72 Answer b: 0.28 Answer c: 0.54 Answer d: We cannot answer because data is missing

In 2016, in France, law enforcement carried out 9.8 million alcohol screening tests with motorists, and $3.1\%$ of these tests were positive. In a given region, on 15 June 2016, a gendarmerie unit conducted screening on 200 motorists. Statement 2: rounding to the nearest hundredth, the probability that, out of the 200 tests, there were strictly more than 5 positive tests, is equal to 0.59. Indicate whether Statement 2 is true or false, justifying your answer.

A general knowledge test consists of a multiple choice questionnaire (MCQ) with twenty questions. For each one, the subject proposes four possible answers, of which only one is correct. For each question, the candidate must necessarily choose a single answer. This person earns one point for each correct answer and loses no points if their answer is wrong.
We consider three candidates:

• Anselme answers completely at random to each of the twenty questions. In other words, for each of the questions, the probability that he answers correctly is equal to $\frac { 1 } { 4 }$;
• Barbara is somewhat better prepared. We consider that for each of the twenty questions, the probability that she answers correctly is $\frac { 1 } { 2 }$;
• Camille does even better: for each of the questions, the probability that she answers correctly is $\frac { 2 } { 3 }$.

1. We denote $X , Y$ and $Z$ the random variables equal to the scores respectively obtained by Anselme, Barbara and Camille. a. What is the probability distribution followed by the random variable $X$? Justify. b. Using a calculator, give the answer rounded to the nearest thousandth of the probability $P ( X \geqslant 10 )$. In the following, we will admit that $P ( Y \geqslant 10 ) \approx 0.588$ and $P ( Z \geqslant 10 ) \approx 0.962$.
2. We randomly choose the copy of one of these three candidates.

We denote $A , B , C$ and $M$ the events:

• $A$: ``the chosen copy is Anselme's'';
• $B$: ``the chosen copy is Barbara's'';
• $C$: ``the chosen copy is Camille's'';
• $M$: ``the chosen copy obtains a score greater than or equal to 10''.

We observe, after correcting it, that the chosen copy obtains a score greater than or equal to 10 out of 20.
What is the probability that it is Barbara's copy? Give the answer rounded to the nearest thousandth of this probability.

Exercise 2 (4 points)
The flu virus affects each year, during the winter period, part of the population of a city. Vaccination against the flu is possible; it must be renewed each year.
Part A
A study conducted in the city's population at the end of the winter period found that:

• $40 \%$ of the population is vaccinated;
• $8 \%$ of vaccinated people contracted the flu;
• $20 \%$ of the population contracted the flu.

A person is chosen at random from the city's population and we consider the events: $V$: ``the person is vaccinated against the flu''; $G$: ``the person contracted the flu''.

1. a. Give the probability of event $G$. b. Reproduce the probability tree below and complete the blanks indicated on four of its branches.
2. Determine the probability that the chosen person contracted the flu and is vaccinated.
3. The chosen person is not vaccinated. Show that the probability that they contracted the flu is equal to 0.28.

Part B
In this part, the probabilities requested will be given to $10 ^ { - 3 }$ near.
A pharmaceutical laboratory conducts a study on vaccination against the flu in this city. After the winter period, $n$ inhabitants of the city are randomly interviewed, assuming that this choice amounts to $n$ successive independent draws with replacement. We assume that the probability that a person chosen at random in the city is vaccinated against the flu is equal to 0.4. Let $X$ be the random variable equal to the number of vaccinated people among the $n$ interviewed.

1. What is the probability distribution followed by the random variable $X$?
2. In this question, we assume that $n = 40$. a. Determine the probability that exactly 15 of the 40 people interviewed are vaccinated. b. Determine the probability that at least half of the people interviewed are vaccinated.
3. A sample of 3750 inhabitants of the city is interviewed, that is, we assume here that $n = 3750$. Let $Z$ be the random variable defined by: $Z = \frac { X - 1500 } { 30 }$. We admit that the probability distribution of the random variable $Z$ can be approximated by the standard normal distribution. Using this approximation, determine the probability that there are between 1450 and 1550 vaccinated individuals in the sample interviewed.

A statistical study established that one in four clients practises surfing.
In a cable car accommodating 80 clients of the resort, the probability rounded to the nearest thousandth that there are exactly 20 clients practising surfing is: a. 0.560 b. 0.25 c. 1 d. 0.103

An urn contains 5 red balls and 3 white balls indistinguishable to the touch.
A ball is drawn from the urn and its colour is noted. This experiment is repeated 4 times, independently, by replacing the ball in the urn each time.
The probability, rounded to the nearest hundredth, of obtaining at least 1 white ball is: Answer A: 0.15 \quad Answer B: 0.63 \quad Answer C: 0.5 \quad Answer D: 0.85

According to a study, regular users of public transport represent $17\%$ of the French population. Among these regular users, $32\%$ are young people aged 18 to 24 years old.
A person is randomly interviewed and we note:

• $R$ the event: ``The person interviewed regularly uses public transport''.
• $J$ the event: ``The person interviewed is aged 18 to 24 years old''.

Part A:

1. Represent the situation using a probability tree, reporting the data from the problem statement.
2. Calculate the probability $P(R \cap J)$.
3. According to this same study, young people aged 18 to 24 represent $11\%$ of the French population. Show that the probability that the person interviewed is a young person aged 18 to 24 who does not regularly use public transport is 0.056 to $10^{-3}$ precision.
4. Deduce the proportion of young people aged 18 to 24 among non-regular users of public transport.

Part B: During a census of the French population, a census taker randomly interviews 50 people in one day about their use of public transport. The French population is large enough to assimilate this census to sampling with replacement. Let $X$ be the random variable counting the number of people regularly using public transport among the 50 people interviewed.

1. Determine, by justifying, the distribution of $X$ and specify its parameters.
2. Calculate $P(X = 5)$ and interpret the result.
3. The census taker indicates that there is more than a $95\%$ chance that, among the 50 people interviewed, fewer than 13 of them regularly use public transport. Is this statement true? Justify your answer.
4. What is the average number of people regularly using public transport among the 50 people interviewed?

This exercise is a multiple choice questionnaire. For each of the following questions, only one of the four proposed answers is correct.
A correct answer earns one point. An incorrect answer, a multiple answer, or the absence of an answer to a question earns or deducts no points.
PART I
In a mail processing centre, a machine is equipped with an automatic optical reader for recognizing postal addresses. This reading system correctly recognizes $97\%$ of addresses; the remaining mail, which will be described as unreadable for the machine, is directed to a centre employee responsible for reading the addresses. This machine has just read nine addresses. We denote by $X$ the random variable that gives the number of unreadable addresses among these nine addresses. We assume that $X$ follows the binomial distribution with parameters $n = 9$ and $p = 0.03$.

1. The probability that none of the nine addresses is unreadable is equal, to the nearest hundredth, to: a. 0 b. 1 c. 0.24 d. 0.76
2. The probability that exactly two of the nine addresses are unreadable for the machine is: a. $\binom{9}{2} \times 0.97^{2} \times 0.03^{7}$ b. $\binom{7}{2} \times 0.97^{2} \times 0.03^{7}$ c. $\binom{9}{2} \times 0.97^{7} \times 0.03^{2}$ d. $\binom{7}{2} \times 0.97^{7} \times 0.03^{2}$
3. The probability that at least one of the nine addresses is unreadable for the machine is: a. $P(X < 1)$ b. $P(X \leqslant 1)$ c. $P(X \geqslant 2)$ d. $1 - P(X = 0)$

PART II
An urn contains 5 green balls and 3 white balls, indistinguishable to the touch. We draw at random successively and without replacement two balls from the urn. We consider the following events:

• $V_{1}$: "the first ball drawn is green";
• $B_{1}$: "the first ball drawn is white";
• $V_{2}$: "the second ball drawn is green";
• $B_{2}$: "the second ball drawn is white".

\setcounter{enumi}{3}
1. The probability of $V_{2}$ given that $V_{1}$ is realized, denoted $P_{V_{1}}\left(V_{2}\right)$, is equal to: a. $\frac{5}{8}$ b. $\frac{4}{7}$ c. $\frac{5}{14}$ d. $\frac{20}{56}$
2. The probability of event $V_{2}$ is equal to: a. $\frac{5}{8}$ b. $\frac{5}{7}$ c. $\frac{3}{28}$ d. $\frac{9}{7}$

Question 3: In an urn there are 6 black balls and 4 red balls. We perform 10 successive random draws with replacement. What is the probability (to $10^{-4}$ near) of obtaining 4 black balls and 6 red balls?

a. 0.1662

b. 0.4

c. 0.1115

d. 0.8886

Exercise 1 Probability
The alarm system of a company operates in such a way that, if a danger presents itself, the alarm activates with a probability of 0.97. The probability that a danger presents itself is 0.01 and the probability that the alarm activates is 0.01465. We denote $A$ the event ``the alarm activates'' and $D$ the event ``a danger presents itself''. We denote $\bar{M}$ the opposite event of an event $M$ and $P(M)$ the probability of the event $M$.

PART A

1. Represent the situation with a weighted tree diagram that will be completed as the exercise progresses.
2. a. Calculate the probability that a danger presents itself and the alarm activates. b. Deduce from this the probability that a danger presents itself given that the alarm activates. Round the result to $10^{-3}$.
3. Show that the probability that the alarm activates given that no danger has presented itself is 0.005.
4. An alarm is considered not to function normally when a danger presents itself and it does not activate, or when no danger presents itself and it activates. Show that the probability that the alarm does not function normally is less than 0.01.

PART B

A factory manufactures alarm systems in large quantities. We successively and randomly select 5 alarm systems from the factory's production. This selection is treated as sampling with replacement. We denote $S$ the event ``the alarm does not function normally'' and we admit that $P(S) = 0.00525$. We consider $X$ the random variable that gives the number of alarm systems not functioning normally among the 5 alarm systems selected. Results should be rounded to $10^{-4}$.

1. Give the probability distribution followed by the random variable $X$ and specify its parameters.
2. Calculate the probability that, in the selected batch, only one alarm system does not function normally.
3. Calculate the probability that, in the selected batch, at least one alarm system does not function normally.

PART C

Let $n$ be a non-zero natural integer. We successively and randomly select $n$ alarm systems. This selection is treated as sampling with replacement. Determine the smallest integer $n$ such that the probability of having, in the selected batch, at least one alarm system that does not function normally is greater than 0.07.

Exercise 1 (7 points) — Main topics covered: Probability
In basketball, there are two types of shots:

• two-point shots: taken near the basket and score two points if successful.
• three-point shots: taken far from the basket and score three points if successful.

Stéphanie is practising shooting. We have the following data:

• One quarter of her shots are two-point shots. Among these, $60\%$ are successful.
• Three quarters of her shots are three-point shots. Among these, $35\%$ are successful.

1. Stéphanie takes a shot. Consider the following events: $D$: ``It is a two-point shot''. $R$: ``the shot is successful''. a. Represent the situation using a probability tree. b. Calculate the probability $p(\bar{D} \cap R)$. c. Prove that the probability that Stéphanie successfully makes a shot is equal to 0.4125. d. Stéphanie successfully makes a shot. Calculate the probability that it is a three-point shot. Round the result to the nearest hundredth.
   
2. Stéphanie now takes a series of 10 three-point shots. Let $X$ be the random variable that counts the number of successful shots. Consider that the shots are independent. Recall that the probability that Stéphanie successfully makes a three-point shot is equal to 0.35. a. Justify that $X$ follows a binomial distribution. Specify its parameters. b. Calculate the expected value of $X$. Interpret the result in the context of the exercise. c. Determine the probability that Stéphanie misses 4 or more shots. Round the result to the nearest hundredth. d. Determine the probability that Stéphanie misses at most 4 shots. Round the result to the nearest hundredth.
   
3. Let $n$ be a non-zero natural number. Stéphanie wishes to take a series of $n$ three-point shots. Consider that the shots are independent. Recall that the probability that she successfully makes a three-point shot is equal to 0.35. Determine the minimum value of $n$ so that the probability that Stéphanie successfully makes at least one shot among the $n$ shots is greater than or equal to 0.99.

Exercise 1 — 6 points

Main topics covered: Probability

At a ski resort, there are two types of passes depending on the skier's age:

• a JUNIOR pass for people under 25 years old;
• a SENIOR pass for others.

Furthermore, a user can choose, in addition to the pass corresponding to their age, the skip-the-line option which allows them to reduce waiting time at the ski lifts. We assume that:

• $20 \%$ of skiers have a JUNIOR pass;
• $80 \%$ of skiers have a SENIOR pass;
• among skiers with a JUNIOR pass, $6 \%$ choose the skip-the-line option;
• among skiers with a SENIOR pass, $12.5 \%$ choose the skip-the-line option.

We interview a skier at random and consider the events:

• $J$ : ``the skier has a JUNIOR pass'';
• $C$ : ``the skier chooses the skip-the-line option''.

The two parts can be worked on independently

Part A

1. Represent the situation with a probability tree.
2. Calculate the probability $P ( J \cap C )$.
3. Prove that the probability that the skier chooses the skip-the-line option is equal to 0.112.
4. The skier has chosen the skip-the-line option. What is the probability that this is a skier with a SENIOR pass? Round the result to $10 ^ { - 3 }$.
5. Is it true that people under twenty-five years old represent less than $15 \%$ of skiers who chose the skip-the-line option? Explain.

Part B

We recall that the probability that a skier chooses the skip-the-line option is equal to 0.112. We consider a sample of 30 skiers chosen at random. Let $X$ be the random variable that counts the number of skiers in the sample who chose the skip-the-line option.

1. We assume that the random variable $X$ follows a binomial distribution. Give the parameters of this distribution.
2. Calculate the probability that at least one of the 30 skiers chose the skip-the-line option. Round the result to $10 ^ { - 3 }$.
3. Calculate the probability that at most one of the 30 skiers chose the skip-the-line option. Round the result to $10 ^ { - 3 }$.
4. Calculate the expected value of the random variable $X$.

Exercise 1 — Theme: Probability Results should be rounded if necessary to $10^{-4}$
A statistical study conducted in a company provides the following information:

• $48\%$ of employees are women. Among them, $16.5\%$ hold a managerial position;
• $52\%$ of employees are men. Among them, $21.5\%$ hold a managerial position.

A person is chosen at random from among the employees. The following events are considered:

• $F$: ``the chosen person is a woman'';
• $C$: ``the chosen person holds a managerial position''.

1. Represent the situation with a probability tree.
2. Calculate the probability that the chosen person is a woman who holds a managerial position.
3. a. Prove that the probability that the chosen person holds a managerial position is equal to 0.191. b. Are the events $F$ and $C$ independent? Justify.
4. Calculate the probability of $F$ given $C$, denoted $P_{C}(F)$. Interpret the result in the context of the exercise.
5. A random sample of 15 employees is chosen. The large number of employees in the company allows this choice to be treated as sampling with replacement. Let $X$ be the random variable giving the number of managers in the sample of 15 employees. Recall that the probability that a randomly chosen employee is a manager is equal to 0.191. a. Justify that $X$ follows a binomial distribution and specify its parameters. b. Calculate the probability that the sample contains at most 1 manager. c. Determine the expected value of the random variable $X$.
6. Let $n$ be a natural number. In this question, consider a sample of $n$ employees. What must be the minimum value of $n$ so that the probability that there is at least one manager in the sample is greater than or equal to 0.99?

Exercise 1 (7 points) -- Probabilities
Among sore throats, one quarter requires taking antibiotics, the others do not. In order to avoid unnecessarily prescribing antibiotics, doctors have a diagnostic test with the following characteristics:

• when the sore throat requires taking antibiotics, the test is positive in $90\%$ of cases;
• when the sore throat does not require taking antibiotics, the test is negative in $95\%$ of cases.

The probabilities requested in the rest of the exercise will be rounded to $10^{-4}$ if necessary.
Part 1
A patient with a sore throat who has undergone the test is chosen at random. Consider the following events:

• $A$: ``the patient has a sore throat requiring taking antibiotics'';
• $T$: ``the test is positive'';
• $\bar{A}$ and $\bar{T}$ are respectively the complementary events of $A$ and $T$.

1. Calculate $P(A \cap T)$. You may use a probability tree.
2. Prove that $P(T) = 0.2625$.
3. A patient with a positive test is chosen. Calculate the probability that they have a sore throat requiring taking antibiotics.
4. a. Among the following events, determine which correspond to an incorrect test result: $A \cap T,\ \bar{A} \cap T,\ A \cap \bar{T},\ \bar{A} \cap \bar{T}$. b. Define the event $E$: ``the test gives an incorrect result''. Prove that $P(E) = 0.0625$.

Part 2
A sample of $n$ patients who have been tested is selected at random. We assume that this sample selection can be treated as sampling with replacement. Let $X$ be the random variable giving the number of patients in this sample with an incorrect test result.

1. Suppose that $n = 50$. a. Justify that the random variable $X$ follows a binomial distribution $\mathscr{B}(n, p)$ with parameters $n = 50$ and $p = 0.0625$. b. Calculate $P(X = 7)$. c. Calculate the probability that there is at least one patient in the sample whose test is incorrect.
2. What is the minimum sample size needed so that $P(X \geqslant 10)$ is greater than $0.95$?

In this part, we model the situation as follows:

• the condition of a scooter is independent of that of the others;
• the probability that a scooter is in good condition is equal to 0.8.

We denote $X$ the random variable which, to a batch of 15 scooters, associates the number of scooters in good condition. Since the number of scooters in the fleet is very large, the sampling of 15 scooters can be assimilated to a draw with replacement.

1. Justify that $X$ follows a binomial distribution and specify the parameters of this distribution.
2. Calculate the probability that all 15 scooters are in good condition.
3. Calculate the probability that at least 10 scooters are in good condition in a batch of 15.
4. We admit that $E(X) = 12$. Interpret the result.

A video game has a large community of online players. Before starting a game, the player must choose between two ``worlds'': either world A or world B. An individual is chosen at random from the community of players. When playing a game, we assume that:

• the probability that the player chooses world A is equal to $\frac{2}{5}$;
• if the player chooses world A, the probability that they win the game is $\frac{7}{10}$;
• the probability that the player wins the game is $\frac{12}{25}$.

We consider the following events:

• A: ``The player chooses world A'';
• B: ``The player chooses world B'';
• G: ``The player wins the game''.

This exercise is a multiple choice questionnaire (5 questions). For each question, only one of the four proposed answers is correct.

1. The probability that the player chooses world A and wins the game is equal to: a. $\frac{7}{10}$ b. $\frac{3}{25}$ c. $\frac{7}{25}$ d. $\frac{24}{125}$
2. The probability $P_{B}(G)$ of event $G$ given that $B$ is realized is equal to: a. $\frac{1}{5}$ b. $\frac{1}{3}$ c. $\frac{7}{15}$ d. $\frac{5}{12}$

In the rest of the exercise, a player plays 10 successive games. This situation is treated as a random draw with replacement. We recall that the probability of winning a game is $\frac{12}{25}$.
3. The probability, rounded to the nearest thousandth, that the player wins exactly 6 games is equal to: a. 0.859 b. 0.671 c. 0.188 d. 0.187
4. We consider a natural number $n$ for which the probability, rounded to the nearest thousandth, that the player wins at most $n$ games is 0.207. Then: a. $n = 2$ b. $n = 3$ c. $n = 4$ d. $n = 5$
5. The probability that the player wins at least one game is equal to: a. $1 - \left(\frac{12}{25}\right)^{10}$ b. $\left(\frac{13}{25}\right)^{10}$ c. $\left(\frac{12}{25}\right)^{10}$ d. $1 - \left(\frac{13}{25}\right)^{10}$

A merchant sells two types of mattresses: SPRING mattresses and FOAM mattresses. We assume that each customer buys only one mattress.
We have the following information:

• $20\%$ of customers buy a SPRING mattress. Among them, $90\%$ are satisfied with their purchase.
• $82\%$ of customers are satisfied with their purchase.

The two parts can be treated independently.
Part A
We randomly select a customer and note the events:

• R: ``the customer buys a SPRING mattress'',
• S: ``the customer is satisfied with their purchase''.

We denote $x = P_{\bar{R}}(S)$, where $P_{\bar{R}}(S)$ denotes the probability of $S$ given that $R$ is not realized.

1. Copy and complete the probability tree below describing the situation.
2. Prove that $x = 0.8$.
3. A customer satisfied with their purchase is selected. What is the probability that they bought a SPRING mattress? Round the result to $10^{-2}$.

Part B

1. We randomly select 5 customers. We consider the random variable $X$ which gives the number of customers satisfied with their purchase among these 5 customers.
   a. We admit that $X$ follows a binomial distribution. Give its parameters.
   b. Determine the probability that at most three customers are satisfied with their purchase. Round the result to $10^{-3}$.
   
2. Let $n$ be a non-zero natural number. We now randomly select $n$ customers. This selection can be treated as a random draw with replacement.
   a. We denote $p_n$ the probability that all $n$ customers are satisfied with their purchase. Prove that $p_n = 0.82^n$.
   b. Determine the natural numbers $n$ such that $p_n < 0.01$. Interpret in the context of the exercise.

1. Between 1998 and 2020, in France 18221965 deliveries were recorded, of which 293898 resulted in the birth of twins and 4921 resulted in the birth of at least three children. a. With a precision of $0.1\%$ calculate, among all recorded deliveries, the percentage of deliveries resulting in the birth of twins over the period 1998-2020. b. Verify that the percentage of deliveries that resulted in the birth of at least three children is less than $0.1\%$.

We then consider that this percentage is negligible. We call an ordinary delivery a delivery resulting in the birth of a single child. We call a double delivery a delivery resulting in the birth of exactly two children. We consider in the rest of the exercise that a delivery is either ordinary or double. The probability of an ordinary delivery is equal to 0.984 and that of a double delivery is then equal to 0.016. The probabilities calculated in the rest will be rounded to the nearest thousandth.
2. We admit that on a given day in a maternity ward, $n$ deliveries are performed. We consider that these $n$ deliveries are independent of each other. We denote $X$ the random variable that gives the number of double deliveries performed that day. a. In the case where $n = 20$, specify the probability distribution followed by the random variable $X$ and calculate the probability that exactly one double delivery is performed. b. By the method of your choice that you will explain, determine the smallest value of $n$ such that $P ( X \geqslant 1 ) \geqslant 0.99$. Interpret the result in the context of the exercise.
3. In this maternity ward, among double births, it is estimated that there are $30\%$ monozygotic twins (called ``identical twins'' which are necessarily of the same sex: two boys or two girls) and therefore $70\%$ dizygotic twins (called ``fraternal twins'', which can be of different sexes: two boys, two girls or one boy and one girl). In the case of double births, we admit that, as for ordinary births, the probability of being a girl at birth is equal to 0.49 and that of being a boy at birth is equal to 0.51. In the case of a double birth of dizygotic twins, we also admit that the sex of the second newborn of the twins is independent of the sex of the first newborn. We randomly choose a double delivery performed in this maternity ward and we consider the following events:

• $M$ : ``the twins are monozygotic'';
• $F _ { 1 }$ : ``the first newborn is a girl'';
• $F _ { 2 }$ : ``the second newborn is a girl''.

We will denote $P ( A )$ the probability of event $A$ and $\bar { A }$ the opposite event of $A$. a. Copy and complete the probability tree. b. Show that the probability that the two newborns are girls is 0.315 07. c. The two newborns are twin girls. Calculate the probability that they are monozygotic.

This exercise is a multiple choice questionnaire. For each question, only one of the four proposed answers is correct. No justification is required.
A wrong answer, multiple answers or the absence of an answer to a question neither awards nor deducts points. The five questions are independent.
A production line produces mechanical parts. It is estimated that $4 \%$ of the parts produced by this line are defective. We randomly choose $n$ parts produced by the production line. The number of parts produced is large enough that this choice can be treated as a draw with replacement. We denote $X$ the random variable equal to the number of defective parts drawn. In the following three questions, we take $n = 50$.

1. What is the probability, rounded to the nearest thousandth, of drawing at least one defective part? a. 1 b. 0,870 c. 0,600 d. 0,599
2. The probability $p ( 3 < X \leqslant 7 )$ is equal to : a. $p ( X \leqslant 7 ) - p ( X > 3 )$ b. $p ( X \leqslant 7 ) - p ( X \leqslant 3 )$ c. $p ( X < 7 ) - p ( X > 3 )$ d. $p ( X < 7 ) - p ( X \geqslant 3 )$
3. What is the smallest natural integer $k$ such that the probability of drawing at most $k$ defective parts is greater than or equal to $95 \%$ ? a. 2 b. 3 c. 4 d. 5

In the following questions, $n$ no longer necessarily equals 50.
4. What is the probability of drawing only defective parts? a. $0,04 ^ { n }$ b. $0,96 ^ { n }$ c. $1 - 0,04 ^ { n }$ d. $1 - 0,96 ^ { n }$
5. Consider the Python function below. What does it return?
\begin{verbatim} def seuil (x) : n=1 while 1-0.96**n a. The smallest number $n$ such that the probability of drawing at least one defective part is greater than or equal to x . b. The smallest number $n$ such that the probability of drawing no defective parts is greater than or equal to x. c. The largest number $n$ such that the probability of drawing only defective parts is greater than or equal to x. d. The largest number $n$ such that the probability of drawing no defective parts is greater than or equal to x.

We randomly choose, independently, $n$ machines from the company, where $n$ denotes a non-zero natural integer. We assimilate this choice to a sampling with replacement, and we denote by $X$ the random variable that associates to each batch of $n$ machines the number of defective machines in this batch. We admit that $X$ follows the binomial distribution with parameters $n$ and $p = 0.082$.
In this question, we take $n = 50$.
The value of the probability $p(X > 2)$, rounded to the nearest thousandth, is: a. 0.136 b. 0.789 c. 0.864 d. 0.924