Exercise 4 -- Candidates who have not followed the specialisation course
Part A
During an evening, a television channel broadcast a match. This channel then offered a programme analysing this match. We have the following information:

• $56\%$ of viewers watched the match;
• one quarter of viewers who watched the match also watched the programme;
• $16.2\%$ of viewers watched the programme.

We randomly interview a viewer. We denote the events:

• $M$: ``the viewer watched the match'';
• $E$: ``the viewer watched the programme''.

We denote by $x$ the probability that a viewer watched the programme given that they did not watch the match.

1. Construct a probability tree illustrating the situation.
2. Determine the probability of $M \cap E$.
3. 1. [a.] Verify that $p ( E ) = 0.44 x + 0.14$.
   2. [b.] Deduce the value of $x$.
4. The interviewed viewer did not watch the programme. What is the probability, rounded to $10 ^ { - 2 }$, that they watched the match?

Part B
This institute decides to model the time spent, in hours, by a viewer watching television on the evening of the match, by a random variable $T$ following the normal distribution with mean $\mu = 1.5$ and standard deviation $\sigma = 0.5$.

1. What is the probability, rounded to $10 ^ { - 3 }$, that a viewer spent between one hour and two hours watching television on the evening of the match?
2. Determine the approximation to $10 ^ { - 2 }$ of the real number $t$ such that $P ( T \geqslant t ) = 0.066$. Interpret the result.

Part C
The lifetime of an individual set-top box, expressed in years, is modelled by a random variable denoted $S$ which follows an exponential distribution with parameter $\lambda$ strictly positive. The probability density function of $S$ is the function $f$ defined on $[ 0 ; + \infty [$ by $$f ( x ) = \lambda \mathrm { e } ^ { - \lambda x }$$ The polling institute has observed that one quarter of the set-top boxes have a lifetime between one and two years. The factory that manufactures the set-top boxes claims that their average lifetime is greater than three years. Is the factory's claim correct? The answer must be justified.

Exercise 4 (5 points) — Candidates who have not followed the specialization course
Each week, a farmer offers for direct sale to each of his customers a basket of fresh products that contains a single bottle of fruit juice. A statistical study carried out gives the following results:

• at the end of the first week, the probability that a customer returns the bottle from his basket is 0.9;
• if the customer returned the bottle from his basket one week, then the probability that he brings back the bottle from the basket the following week is 0.95;
• if the customer did not return the bottle from his basket one week, then the probability that he brings back the bottle from the basket the following week is 0.2.

A customer is chosen at random from the farmer's clientele. For any non-zero natural number $n$, we denote by $R_n$ the event ``the customer returns the bottle from his basket in the $n$-th week''.
1. a. Model the situation studied for the first two weeks using a weighted tree that will involve the events $R_1$ and $R_2$.
b. Determine the probability that the customer returns the bottles from the baskets of the first and second weeks.
c. Show that the probability that the customer returns the bottle from the basket of the second week is equal to 0.875.
d. Given that the customer returned the bottle from his basket in the second week, what is the probability that he did not return the bottle from his basket in the first week? Round the result to $10^{-3}$.
2. For any non-zero natural number $n$, we denote by $r_n$ the probability that the customer returns the bottle from the basket in the $n$-th week. We then have $r_n = p(R_n)$.
a. Copy and complete the weighted tree (no justification is required).
b. Justify that for any non-zero natural number $n$, $r_{n+1} = 0.75r_n + 0.2$.
c. Prove that for any non-zero natural number $n$, $r_n = 0.1 \times 0.75^{n-1} + 0.8$.
d. Calculate the limit of the sequence $(r_n)$. Interpret the result in the context of the exercise.

Part A

Louise drives to work with her car. Her colleague Zoé does not own a car. Each morning, Louise therefore offers to give Zoé a ride. Whatever Zoé's answer, Louise offers to drive her back in the evening. We consider a given day. We have the following information:

• the probability that Louise drives Zoé in the morning is 0.55;
• if Louise drove Zoé in the morning, the probability that she drives her back in the evening is 0.7;
• if Louise did not drive Zoé in the morning, the probability that she drives her back in the evening is 0.24.

We denote $M$ and $S$ the following events:

• $M$: ``Louise drives Zoé in the morning'';
• S: ``Louise drives Zoé back in the evening''.

1. Construct a probability tree representing the situation.
2. Calculate $P ( M \cap S )$. Translate this result with a sentence.
3. Prove that the probability of event S is equal to 0.493.
4. We know that Louise drove Zoé back in the evening. What is the probability that Louise drove her in the morning?

Consider $T$ the random variable following the normal distribution with mean $\mu = 60$ and standard deviation $\sigma = 6$.
The probability $P _ { ( T > 60 ) } ( T > 72 )$ rounded to the nearest thousandth is: Answer A: 0.954 Answer B: 1 Answer C: 0.023 Answer D: 0.046

A pharmaceutical laboratory has just developed a new anti-doping test.

Part A

A study on this new test gives the following results:

• if an athlete is doped, the probability that the test result is positive is 0.98 (test sensitivity);
• if an athlete is not doped, the probability that the test result is negative is 0.995 (test specificity).

The test is administered to an athlete selected at random from among the participants in an athletics competition. We denote by $D$ the event ``the athlete is doped'' and $T$ the event ``the test is positive''. We assume that the probability of event $D$ is equal to 0.08.

1. Represent the situation in the form of a probability tree.
2. Prove that $P ( T ) = 0.083$.
3. a. Given that an athlete presents a positive test, what is the probability that he is doped? b. The laboratory decides to market the test if the probability of the event ``an athlete presenting a positive test is doped'' is greater than or equal to 0.95. Will the test proposed by the laboratory be marketed? Justify.

Part B

In a sporting competition, we assume that the probability that a tested athlete presents a positive test is 0.103.

1. In this question 1., we assume that the organizers decide to test 5 athletes selected at random from among the athletes in this competition. We denote by $X$ the random variable equal to the number of athletes presenting a positive test among the 5 tested athletes. a. Give the distribution followed by the random variable $X$. Specify its parameters. b. Calculate the expectation $E ( X )$ and interpret the result in the context of the exercise. c. What is the probability that at least one of the 5 tested athletes presents a positive test?
2. How many athletes must be tested at minimum so that the probability of the event ``at least one tested athlete presents a positive test'' is greater than or equal to 0.75? Justify.

A company receives numerous emails daily. Among these emails, $8\%$ are ``spam'', that is, emails with advertising or malicious intent, which it is desirable not to open. An email received by the company is chosen at random. The properties of the email software used in the company allow us to state that:

• The probability that the chosen email is classified as ``undesirable'' given that it is spam is equal to 0.9.
• The probability that the chosen email is classified as ``undesirable'' given that it is not spam is equal to 0.01.

We denote:

• S the event ``the chosen email is spam'';
• I the event ``the chosen email is classified as undesirable by the email software''.
• $\bar{S}$ and $\bar{I}$ the complementary events of $S$ and $I$ respectively.

1. Model the situation studied using a probability tree, on which the probabilities associated with each branch should appear.
2. a. Prove that the probability that the chosen email is a spam message and is classified as undesirable is equal to 0.072. b. Calculate the probability that the chosen message is classified as undesirable. c. The chosen message is classified as undesirable. What is the probability that it is actually a spam message? Give the answer rounded to the nearest hundredth.
3. A random sample of 50 emails is chosen from those received by the company. It is assumed that this choice amounts to a random draw with replacement of 50 emails from the set of all emails received by the company. Let $Z$ be the random variable counting the number of spam emails among the 50 chosen. a. What is the probability distribution followed by the random variable $Z$, and what are its parameters? b. What is the probability that, among the 50 chosen emails, at least two are spam? Give the answer rounded to the nearest hundredth.

In a statistics school, after reviewing candidate files, recruitment is done in two ways:

• $10\%$ of candidates are selected based on their file. These candidates must then take an oral examination after which $60\%$ of them are finally admitted to the school.
• Candidates who were not selected based on their file take a written examination after which $20\%$ of them are admitted to the school.

Part 1
A candidate for this recruitment competition is chosen at random. We denote:

• $D$ the event ``the candidate was selected based on their file'';
• $A$ the event ``the candidate was admitted to the school'';
• $\bar{D}$ and $\bar{A}$ the complementary events of events $D$ and $A$ respectively.

1. Represent the situation with a probability tree.
2. Calculate the probability that the candidate is selected based on their file and admitted to the school.
3. Show that the probability of event $A$ is equal to 0.24.
4. A candidate admitted to the school is chosen at random. What is the probability that their file was not selected?

Part 2

1. We assume that the probability for a candidate to be admitted to the school is equal to 0.24.
   We consider a sample of seven candidates chosen at random, treating this choice as a random draw with replacement. We denote by $X$ the random variable counting the candidates admitted to the school among the seven drawn. a. We assume that the random variable $X$ follows a binomial distribution. What are the parameters of this distribution? b. Calculate the probability that only one of the seven candidates drawn is admitted to the school. Give an answer rounded to the nearest hundredth. c. Calculate the probability that at least two of the seven candidates drawn are admitted to this school. Give an answer rounded to the nearest hundredth.
2. A secondary school presents $n$ candidates for recruitment in this school, where $n$ is a non-zero natural number. We assume that the probability for any candidate from the secondary school to be admitted to the school is equal to 0.24 and that the results of the candidates are independent of each other. a. Give the expression, as a function of $n$, of the probability that no candidate from this secondary school is admitted to the school. b. From what value of the integer $n$ is the probability that at least one student from this secondary school is admitted to the school greater than or equal to 0.99?

A manufacturing line produces mechanical parts. It is estimated that $5 \%$ of the parts produced by this line are defective.
An engineer has developed a test to apply to the parts. This test has two possible results: ``positive'' or ``negative''. This test is applied to a part chosen at random from the production of the line. We denote $p ( E )$ the probability of an event $E$. We consider the following events:

• $D$: ``the part is defective'';
• T: ``the part shows a positive test'';
• $\bar { D }$ and $\bar { T }$ denote respectively the complementary events of $D$ and $T$.

Given the characteristics of the test, we know that:

• The probability that a part shows a positive test given that it is defective is equal to 0.98;
• the probability that a part shows a negative test given that it is not defective is equal to 0.97.

PART I

1. Represent the situation using a probability tree.
2. a. Determine the probability that a part chosen at random from the production line is defective and shows a positive test. b. Prove that: $p ( T ) = 0.0775$.
3. The positive predictive value of the test is called the probability that a part is defective given that the test is positive. A test is considered effective if it has a positive predictive value greater than 0.95.

Calculate the positive predictive value of this test and specify whether it is effective.

PART II

A sample of 20 parts is chosen from the production line, treating this choice as a draw with replacement. Let $X$ be the random variable that gives the number of defective parts in this sample. Recall that: $p ( D ) = 0.05$.

1. Justify that $X$ follows a binomial distribution and determine the parameters of this distribution.
2. Calculate the probability that this sample contains at least one defective part.

Give a result rounded to the nearest hundredth.
3. Calculate the expected value of the random variable $X$ and interpret the result obtained.

In this exercise, the results of the probabilities requested will be, if necessary, rounded to the nearest thousandth.
Feline leukaemia is a disease affecting cats; it is caused by a virus. In a large veterinary centre, the proportion of cats carrying the disease is estimated at $40\%$. A screening test for the disease is carried out among the cats present in this veterinary centre. This test has the following characteristics.

• When the cat carries the disease, its test is positive in $90\%$ of cases.
• When the cat does not carry the disease, its test is negative in $85\%$ of cases.

A cat is chosen at random from the veterinary centre and the following events are considered:

• $M$: ``The cat carries the disease'';
• $T$: ``The cat's test is positive'';
• $\bar{M}$ and $\bar{T}$ denote the complementary events of events $M$ and $T$ respectively.

1. a. Represent the situation with a probability tree. b. Calculate the probability that the cat carries the disease and that its test is positive. c. Show that the probability that the cat's test is positive is equal to 0.45. d. A cat is chosen from among those whose test is positive. Calculate the probability that it carries the disease.
2. A sample of 20 cats is chosen at random from the veterinary centre. It is assumed that this choice can be treated as sampling with replacement.

Let $X$ be the random variable giving the number of cats presenting a positive test in the chosen sample. a. Determine, by justifying, the distribution followed by the random variable $X$. b. Calculate the probability that there are exactly 5 cats with a positive test in the sample. c. Calculate the probability that there are at most 8 cats with a positive test in the sample. d. Determine the expected value of the random variable $X$ and interpret the result in the context of the exercise.
3. In this question, a sample of $n$ cats is chosen from the centre, which is again treated as sampling with replacement. Let $p_n$ be the probability that there is at least one cat presenting a positive test in this sample. a. Show that $p_n = 1 - 0{,}55^n$. b. Describe the role of the program below written in Python language, in which the variable $n$ is a natural integer and the variable $P$ is a real number. \begin{verbatim} def seuil() : n = 0 P = 0 while P < 0,99 : n = n + 1 P = 1 - 0,55**n return n \end{verbatim} c. Determine, by specifying the method used, the value returned by this program.

A test is developed to detect a disease in a country. According to the health authorities of this country, $7\%$ of the inhabitants are infected with this disease. Among infected individuals, $20\%$ test negative. Among healthy individuals, $1\%$ test positive.
A person is chosen at random from the population. We denote:

• $M$ the event: ``the person is infected with the disease'';
• $T$ the event: ``the test is positive''.

1. Construct a probability tree modelling the proposed situation.
2. a. What is the probability that the person is infected with the disease and that their test is positive? b. Show that the probability that their test is positive is 0.0653.
3. It is known that the test of the chosen person is positive. What is the probability that they are infected? Give the result as an approximation to $10 ^ { - 2 }$ near.
4. Ten people are chosen at random from the population. The size of the population of this country allows us to treat this sample as a draw with replacement. Let $X$ be the random variable that counts the number of individuals with a positive test among the ten people. a. What is the probability distribution followed by $X$? Specify its parameters. b. Determine the probability that exactly two people have a positive test. Give the result as an approximation to $10 ^ { - 2 }$ near.
5. Determine the minimum number of people to test in this country so that the probability that at least one of these people has a positive test is greater than $99\%$.

A hotel located near a prehistoric tourism site offers two visits in the surrounding area, one to a museum and one to a cave.
A study showed that $70\%$ of the hotel's clients visit the museum. Furthermore, among clients visiting the museum, $60\%$ visit the cave. The study also shows that $6\%$ of the hotel's clients make no visits. We randomly question a hotel client and note:

• $M$ the event: ``the client visits the museum'';
• $G$ the event: ``the client visits the cave''.

We denote by $\bar { M }$ the complementary event of $M$, $\bar { G }$ the complementary event of $G$, and for any event $E$, we denote by $p ( E )$ the probability of $E$. Thus, according to the problem statement, we have: $p ( \bar { M } \cap \bar { G } ) = 0.06$.

1. a. Verify that $p _ { \bar { M } } ( \bar { G } ) = 0.2$, where $p _ { \bar { M } } ( \bar { G } )$ denotes the probability that the questioned client does not visit the cave given that he does not visit the museum. b. The weighted tree opposite models the situation. Copy and complete this tree by indicating on each branch the associated probability. c. What is the probability of the event ``the client visits the cave and does not visit the museum''? d. Show that $p ( G ) = 0.66$.
2. The hotel manager claims that among clients who visit the cave, more than half also visit the museum. Is this claim correct?
3. The prices for visits are as follows:
   • museum visit: 12 euros;
   • cave visit: 5 euros.
   We consider the random variable $T$ which models the amount spent by a hotel client for these visits. a. Give the probability distribution of $T$. Present the results in the form of a table. b. Calculate the mathematical expectation of $T$. c. For profitability reasons, the hotel manager estimates that the average amount of visit revenues must be greater than 700 euros per day. Determine the average number of clients per day needed to achieve this objective.
4. To increase revenues, the manager wishes the expectation of the random variable modeling the amount spent by a hotel client for these visits to increase to 15 euros, without changing the museum visit price which remains at 12 euros. What price should be set for the cave visit to achieve this objective? (We will assume that the increase in the cave entrance price does not change the frequency of visits to the two sites).
5. We randomly choose 100 hotel clients, treating this choice as a draw with replacement. What is the probability that at least three-quarters of these clients visited the cave during their stay at the hotel? Give a value of the result to $10 ^ { - 3 }$ near.

Every day he works, Paul must go to the station to reach his workplace by train. To do this, he takes his bicycle two times out of three and, if he does not take his bicycle, he takes his car.

1. When he takes his bicycle to reach the station, Paul misses the train only once in 50, whereas when he takes his car to reach the station, Paul misses his train once in 10. We consider a random day on which Paul will be at the station to catch the train that will take him to work. We denote:
   • V the event ``Paul takes his bicycle to reach the station'';
   • R the event ``Paul misses his train''. a. Draw a weighted tree summarizing the situation. b. Show that the probability that Paul misses his train is equal to $\frac { 7 } { 150 }$. c. Paul has missed his train. Determine the exact value of the probability that he took his bicycle to reach the station.
   
   
2. A random month is chosen during which Paul went to the station 20 days to reach his workplace according to the procedures described in the preamble. We assume that, for each of these 20 days, the choice between bicycle and car is independent of the choices on other days. We denote $X$ the random variable giving the number of days Paul takes his bicycle over these 20 days. a. Determine the distribution followed by the random variable $X$. Specify its parameters. b. What is the probability that Paul takes his bicycle exactly 10 days out of these 20 days to reach the station? Round the probability sought to $10 ^ { - 3 }$. c. What is the probability that Paul takes his bicycle at least 10 days out of these 20 days to reach the station? Round the probability sought to $10 ^ { - 3 }$. d. On average, how many days over a randomly chosen period of 20 days to reach the station does Paul take his bicycle? Round the answer to the nearest integer.
   
3. In the case where Paul goes to the station by car, we denote $T$ the random variable giving the travel time needed to reach the station. The duration of the journey is given in minutes, rounded to the minute. The probability distribution of $T$ is given by the table below:

$k$ (in minutes)	10	11	12	13	14	15	16	17	18
$P ( T = k )$	0,14	0,13	0,13	0,12	0,12	0,11	0,10	0,08	0,07

Determine the expected value of the random variable $T$ and interpret this value in the context of the exercise.

Exercise 1 (7 points) Theme: probabilities, sequences In a tourist region, a company offers a bicycle rental service for the day. The company has two distinct rental points, point A and point B. Bicycles can be borrowed and returned indifferently at either of the two rental points. It is assumed that the total number of bicycles is constant and that every morning, when the service opens, each bicycle is at point A or point B. According to a statistical study:

• If a bicycle is at point A one morning, the probability that it is at point A the next morning is equal to 0.84;
• If a bicycle is at point B one morning, the probability that it is at point B the next morning is equal to 0.76.

When the service opens on the first morning, the company has placed half of its bicycles at point A and the other half at point B. We consider a bicycle from the company chosen at random. For every positive integer $n$, the following events are defined:

• $A _ { n }$ : ``the bicycle is at point A on the $n$-th morning''
• $B _ { n }$ : ``the bicycle is at point B on the $n$-th morning''.

For every positive integer $n$, we denote by $a _ { n }$ the probability of event $A _ { n }$ and by $b _ { n }$ the probability of event $B _ { n }$. Thus $a _ { 1 } = 0.5$ and $b _ { 1 } = 0.5$.

1. Copy and complete the weighted tree that models the situation for the first two mornings.
2. a. Calculate $a _ { 2 }$. b. The bicycle is at point A on the second morning. Calculate the probability that it was at point B on the first morning. The probability will be rounded to the nearest thousandth.
3. a. Copy and complete the weighted tree that models the situation for the $n$-th and $(n + 1)$-th mornings. b. Justify that for every positive integer $n$, $a _ { n + 1 } = 0.6 a _ { n } + 0.24$.
4. Show by induction that, for every positive integer $n$, $a _ { n } = 0.6 - 0.1 \times 0.6 ^ { n - 1 }$.
5. Determine the limit of the sequence $(a _ { n })$ and interpret this limit in the context of the exercise.
6. Determine the smallest positive integer $n$ such that $a _ { n } \geqslant 0.599$ and interpret the result obtained in the context of the exercise.

Exercise 1: Probability
A company manufactures components for the automotive industry. These components are designed on three assembly lines numbered 1 to 3.

• Half of the components are designed on line $\mathrm{n}^{\circ}1$;
• $30\%$ of the components are designed on line $\mathrm{n}^{\circ}2$;
• the remaining components are designed on line $\mathrm{n}^{\circ}3$.

At the end of the manufacturing process, it appears that $1\%$ of the parts from line $\mathrm{n}^{\circ}1$ have a defect, as do $0.5\%$ of the parts from line $\mathrm{n}^{\circ}2$ and $4\%$ of the parts from line $\mathrm{n}^{\circ}3$. One of these components is randomly selected. We denote:

• $C_1$ the event ``the component comes from line $\mathrm{n}^{\circ}1$'';
• $C_2$ the event ``the component comes from line $\mathrm{n}^{\circ}2$'';
• $C_3$ the event ``the component comes from line $\mathrm{n}^{\circ}3$'';
• $D$ the event ``the component is defective'' and $\bar{D}$ its complementary event.

Throughout the exercise, probability calculations will be given as exact decimal values or rounded to $10^{-4}$ if necessary.
PART A

1. Represent this situation with a probability tree.
2. Calculate the probability that the selected component comes from line $\mathrm{n}^{\circ}3$ and is defective.
3. Show that the probability of event $D$ is $P(D) = 0.0145$.
4. Calculate the probability that a defective component comes from line $\mathrm{n}^{\circ}3$.

PART B
The company decides to package the produced components by forming batches of $n$ units. We denote $X$ the random variable which, to each batch of $n$ units, associates the number of defective components in this batch. Given the company's production and packaging methods, we can consider that $X$ follows the binomial distribution with parameters $n$ and $p = 0.0145$.

1. In this question, the batches contain 20 units. We set $n = 20$. a. Calculate the probability that a batch contains exactly three defective components. b. Calculate the probability that a batch contains no defective components. Deduce the probability that a batch contains at least one defective component.
2. The company director wishes the probability of having no defective components in a batch of $n$ components to be greater than 0.85. He proposes to form batches of at most 11 components. Is he correct? Justify your answer.

PART C
The manufacturing costs of the components of this company are 15 euros if they come from assembly line $\mathrm{n}^{\circ}1$, 12 euros if they come from assembly line $\mathrm{n}^{\circ}2$ and 9 euros if they come from assembly line $\mathrm{n}^{\circ}3$. Calculate the average manufacturing cost of a component for this company.

During a fair, a game organizer has, on one hand, a wheel with four white squares and eight red squares and, on the other hand, a bag containing five tokens bearing the numbers $1, 2, 3, 4$ and 5. The game consists of spinning the wheel, each square having equal probability of being obtained, then extracting one or two tokens from the bag according to the following rule:

• if the square obtained by the wheel is white, then the player extracts one token from the bag;
• if the square obtained by the wheel is red, then the player extracts successively and without replacement two tokens from the bag.

The player wins if the token(s) drawn all bear an odd number.

1. A player plays one game and we denote by $B$ the event ``the square obtained is white'', $R$ the event ``the square obtained is red'' and $G$ the event ``the player wins the game''. a. Give the value of the conditional probability $P _ { B } ( G )$. b. It is admitted that the probability of drawing successively and without replacement two odd tokens is equal to 0.3. Copy and complete the following probability tree.
2. a. Show that $P ( G ) = 0.4$. b. A player wins the game. What is the probability that he obtained a white square by spinning the wheel?
3. Are the events $B$ and $G$ independent? Justify.
4. The same player plays ten games. The tokens drawn are returned to the bag after each game. We denote by $X$ the random variable equal to the number of games won. a. Explain why $X$ follows a binomial distribution and specify its parameters. b. Calculate the probability, rounded to $10 ^ { - 3 }$, that the player wins exactly three games out of the ten games played. c. Calculate $P ( X \geqslant 4 )$ rounded to $10 ^ { - 3 }$. Give an interpretation of the result obtained.
5. A player plays $n$ games and we denote by $p _ { n }$ the probability of the event ``the player wins at least one game''. a. Show that $p _ { n } = 1 - 0.6 ^ { n }$. b. Determine the smallest value of the integer $n$ for which the probability of winning at least one game is greater than or equal to 0.99.

Exercise 1 (7 points) Theme: probability The coyote is a wild animal close to the wolf, which lives in North America. In the state of Oklahoma, in the United States, $70\%$ of coyotes are affected by a disease called ehrlichiosis. There is a test that helps detect this disease. When this test is applied to a coyote, its result is either positive or negative, and we know that:

• If the coyote is sick, the test is positive in $97\%$ of cases.
• If the coyote is not sick, the test is negative in $95\%$ of cases.

Part A Veterinarians capture a coyote from Oklahoma at random and perform a test for ehrlichiosis on it. We consider the following events:

• $M$: ``the coyote is sick'';
• $T$: ``the coyote's test is positive''.

We denote $\bar{M}$ and $\bar{T}$ respectively the complementary events of $M$ and $T$.

1. Copy and complete the probability tree below that models the situation.
2. Determine the probability that the coyote is sick and that its test is positive.
3. Prove that the probability of $T$ is equal to 0.694.
4. The ``positive predictive value of the test'' is called the probability that the coyote is actually sick given that its test is positive. Calculate the positive predictive value of the test. Round the result to the nearest thousandth.
5. a. By analogy with the previous question, propose a definition of the ``negative predictive value of the test'' and calculate this value rounding to the nearest thousandth. b. Compare the positive and negative predictive values of the test, and interpret.

Part B Recall that the probability that a randomly captured coyote has a positive test is 0.694.

1. When five coyotes are captured at random, this choice is treated as sampling with replacement. We denote $X$ the random variable that associates to a sample of five randomly captured coyotes the number of coyotes in this sample having a positive test. a. What is the probability distribution followed by $X$? Justify and specify its parameters. b. Calculate the probability that in a sample of five randomly captured coyotes, only one has a positive test. Round the result to the nearest hundredth. c. A veterinarian claims that there is more than a one in two chance that at least four out of five coyotes have a positive test: is this claim true? Justify your answer.
2. To test medications, veterinarians need to have a coyote with a positive test. How many coyotes must they capture so that the probability that at least one of them has a positive test is greater than 0.99?

According to the health authorities of a country, $7\%$ of the inhabitants are affected by a certain disease. In this country, a test is developed to detect this disease. This test has the following characteristics:

• For sick individuals, the test gives a negative result in $20\%$ of cases;
• For healthy individuals, the test gives a positive result in $1\%$ of cases.

A person is chosen at random from the population and tested. Consider the following events:

• $M$ ``the person is sick'';
• $T$ ``the test is positive''.

1. Calculate the probability of the event $M \cap T$. You may use a probability tree.
2. Prove that the probability that the test of the randomly chosen person is positive is $0.0653$.
3. In the context of disease screening, is it more relevant to know $P_M(T)$ or $P_T(M)$?
4. In this question, consider that the randomly chosen person had a positive test. What is the probability that they are sick? Round the result to $10^{-2}$ near.
5. People are chosen at random from the population. The size of the population of this country allows us to treat this sampling as drawing with replacement. Let $X$ be the random variable that gives the number of individuals with a positive test among 10 people. a. Specify the nature and parameters of the probability distribution followed by $X$. b. Determine the probability that exactly two people have a positive test. Round the result to $10^{-2}$ near.
6. Determine the minimum number of people to test in this country so that the probability that at least one of them has a positive test is greater than $99\%$.

Customs authorities are interested in imports of headphones bearing the logo of a certain brand. Customs seizures allow them to estimate that:

• $20 \%$ of headphones bearing this brand's logo are counterfeits;
• $2 \%$ of non-counterfeit headphones have a design defect;
• $10 \%$ of counterfeit headphones have a design defect.

The fraud agency randomly orders a headphone displaying the brand's logo from an internet site. Consider the following events:

• C: ``the headphone is counterfeit'';
• $D$: ``the headphone has a design defect'';
• $\bar { C }$ and $\bar { D }$ denote respectively the complementary events of $C$ and $D$.

Throughout the exercise, probabilities will be rounded to $10 ^ { - 3 }$ if necessary.

Part 1

1. Calculate $P ( C \cap D )$. You may use a probability tree.
2. Prove that $P ( D ) = 0,036$.
3. The headphone has a defect. What is the probability that it is counterfeit?

Part 2

We order $n$ headphones bearing this brand's logo. We treat this experiment as a random draw with replacement. Let $X$ be the random variable giving the number of headphones with a design defect in this batch.

1. In this question, $n = 35$. a. Justify that $X$ follows a binomial distribution $\mathscr { B } ( n , p )$ where $n = 35$ and $p = 0,036$. b. Calculate the probability that among the ordered headphones, exactly one has a design defect. c. Calculate $P ( X \leqslant 1 )$.
2. In this question, $n$ is not fixed. What is the minimum number of headphones to order so that the probability that at least one headphone has a defect is greater than 0.99?

Part 1

Julien must take the plane; he planned to take the bus to get to the airport. If he takes the 8 o'clock bus, he is sure to be at the airport in time for his flight. On the other hand, the next bus would not allow him to arrive at the airport in time. Julien left late from his apartment and the probability that he misses his bus is 0.8. If he misses his bus, he goes to the airport by taking a private car company; he then has a probability of 0.5 of being on time at the airport. We denote:

• $B$ the event: ``Julien manages to take his bus'';
• $V$ the event: ``Julien is on time at the airport for his flight''.

1. Give the value of $P _ { B } ( V )$.
2. Represent the situation with a probability tree.
3. Show that $P ( V ) = 0.6$.
4. If Julien is on time at the airport for his flight, what is the probability that he arrived at the airport by bus? Justify.

Part 2

Airlines sell more tickets than there are seats on planes because some passengers do not show up for boarding on the flight they have booked. This practice is called overbooking. Based on statistics from previous flights, the airline estimates that each passenger has a 5\% chance of not showing up for boarding. Consider a flight on a plane with 200 seats for which 206 tickets have been sold. We assume that the presence at boarding of each passenger is independent of other passengers and we call $X$ the random variable that counts the number of passengers showing up for boarding.

1. Justify that $X$ follows a binomial distribution and specify its parameters.
2. On average, how many passengers will show up for boarding?
3. Calculate the probability that 201 passengers show up for boarding. The result should be rounded to $10 ^ { - 3 }$.
4. Calculate $P ( X \leqslant 200 )$, the result should be rounded to $10 ^ { - 3 }$. Interpret this result in the context of the exercise.
5. The airline sells each ticket for 250 euros.

If more than 200 passengers show up for boarding, the airline must refund the plane ticket and pay a penalty of 600 euros to each affected passenger. We call: $Y$ the random variable equal to the number of passengers who cannot board despite having purchased a ticket; $C$ the random variable that totals the revenue of the airline on this flight.
We admit that $Y$ follows the probability distribution given by the following table:

$y _ { i }$	0	1	2	3	4	5	6
$P \left( Y = y _ { i } \right)$	0,94775	0,03063	0,01441	0,00539	0,00151	0,00028

a. Complete the probability distribution given above by calculating $P ( Y = 6 )$. b. Justify that: $C = 51500 - 850 Y$. c. Give the probability distribution of the random variable $C$ in the form of a table. Calculate the expected value of the random variable $C$ to the nearest euro. d. Compare the revenue obtained by selling exactly 200 tickets and the average revenue obtained by practicing overbooking.

Exercise 3 (7 points) The director of a large company proposed a training course to all its employees on the use of new software. This course was followed by $25\%$ of employees.

1. In this company, $52\%$ of employees are women, of whom $40\%$ followed the course.

A random employee of the company is questioned and we consider the events:

• $F$: ``the employee questioned is a woman'',
• $S$: ``the employee questioned followed the course''. $\bar{F}$ and $\bar{S}$ denote respectively the complementary events of events $F$ and $S$. a. Give the probability of event $S$. b. Copy and complete the blanks of the probability tree below on the four indicated branches. c. Demonstrate that the probability that the person questioned is a woman who followed the course is equal to 0.208. d. Given that the person questioned followed the course, what is the probability that it is a woman? e. The director claims that, among the male employees of the company, fewer than $10\%$ followed the course. Justify the director's claim.

1. We denote by $X$ the random variable that associates to a sample of 20 employees of this company chosen at random the number of employees in this sample who followed the course. We assume that the number of employees in the company is sufficiently large to assimilate this choice to sampling with replacement. a. Determine, by justifying, the probability distribution followed by the random variable $X$. b. Determine, to $10^{-3}$ near, the probability that 5 employees in a sample of 20 followed the course. c. The program below, written in Python language, uses the function binomial$(i, n, p)$ created for this purpose which returns the value of the probability $P(X = i)$ in the case where the random variable $X$ follows a binomial distribution with parameters $n$ and $p$. \begin{verbatim} def proba(k) : P=0 for i in range(0,k+1) : P=P+binomiale(i,20,0.25) return P \end{verbatim} Determine, to $10^{-3}$ near, the value returned by this program when proba(5) is entered in the Python console. Interpret this value in the context of the exercise. d. Determine, to $10^{-3}$ near, the probability that at least 6 employees in a sample of 20 followed the course.
2. This question is independent of questions 1 and 2. To encourage employees to follow the course, the company had decided to increase the salaries of employees who followed the course by $5\%$, compared to $2\%$ increase for employees who did not follow the course. What is the average percentage increase in salaries for this company under these conditions?

Consider a binary communication system transmitting 0s and 1s. Each 0 or 1 is called a bit. Due to interference, there may be transmission errors: a 0 can be received as a 1 and, likewise, a 1 can be received as a 0. For a bit chosen at random in the message, we note the events:

• $E _ { 0 }$: ``the bit sent is a 0'';
• $E _ { 1 }$: ``the bit sent is a 1'';
• $R _ { 0 }$: ``the bit received is a 0''
• $R _ { 1 }$: ``the bit received is a 1''.

We know that: $p \left( E _ { 0 } \right) = 0{,}4 ; \quad p _ { E _ { 0 } } \left( R _ { 1 } \right) = 0{,}01 ; \quad p _ { E _ { 1 } } \left( R _ { 0 } \right) = 0{,}02$. Recall that the conditional probability of $A$ given $B$ is denoted $p _ { B } ( A )$.

1. The probability that the bit sent is a 0 and the bit received is a 0 is equal to: a. 0,99 b. 0,396 c. 0,01 d. 0,4
2. The probability $p \left( R _ { 0 } \right)$ is equal to: a. 0,99 b. 0,02 c. 0,408 d. 0,931
3. A value, approximated to the nearest thousandth, of the probability $p _ { R _ { 1 } } \left( E _ { 0 } \right)$ is equal to: a. 0,004 b. 0,001 c. 0,007 d. 0,010
4. The probability of the event ``there is a transmission error'' is equal to: a. 0,03 b. 0,016 c. 0,16 d. 0,015

A message of length eight bits is called a byte. It is admitted that the probability that a byte is transmitted without error is equal to 0,88.

1. 10 bytes are transmitted successively in an independent manner.
   The probability, to $10 ^ { - 3 }$ near, that exactly 7 bytes are transmitted without error is equal to: a. 0,915 b. 0,109 c. 0,976 d. 0,085
2. 10 bytes are transmitted successively in an independent manner.
   The probability that at least 1 byte is transmitted without error is equal to: a. $1 - 0{,}12 ^ { 10 }$ b. $0{,}12 ^ { 10 }$ c. $0{,}88 ^ { 10 }$ d. $1 - 0{,}88 ^ { 10 }$
3. Let $N$ be a natural integer. $N$ bytes are transmitted successively in an independent manner. Let $N _ { 0 }$ be the largest value of $N$ for which the probability that all $N$ bytes are transmitted without error is greater than or equal to 0,1. We can affirm that: a. $N _ { 0 } = 17$ b. $N _ { 0 } = 18$ c. $N _ { 0 } = 19$ d. $N _ { 0 } = 20$

4 points Parts $\boldsymbol{A}$ and $\boldsymbol{B}$ are independent. The requested probabilities will be given to $10^{-3}$ near. To help detect certain allergies, a blood test can be performed whose result is either positive or negative. In a population, this test gives the following results:

• If an individual is allergic, the test is positive in $97\%$ of cases;
• If an individual is not allergic, the test is negative in $95{,}7\%$ of cases.

Furthermore, $20\%$ of individuals in the concerned population have a positive test. We randomly choose an individual from the population, and we denote:

• $A$ the event ``the individual is allergic'';
• $T$ the event ``the individual has a positive test''.

We denote by $\bar{A}$ and $\bar{T}$ the complementary events of $A$ and $T$. We also call $x$ the probability of event $A$: $x = p(A)$.
Part A

1. Reproduce and complete the tree describing the situation, indicating on each branch the corresponding probability.
2. a. Prove the equality: $p(T) = 0{,}927x + 0{,}043$. b. Deduce the probability that the chosen individual is allergic.
3. Justify by a calculation the following statement: ``If the test of an individual chosen at random is positive, there is more than $80\%$ chance that this individual is allergic''.

Part B
A survey on allergies is conducted in a city by interviewing 150 randomly chosen residents, and we assume that this choice amounts to successive independent draws with replacement. We know that the probability that a randomly chosen resident in this city is allergic is equal to $0{,}08$. We denote by $X$ the random variable that associates to a sample of 150 randomly chosen residents the number of allergic people in this sample.

1. What is the probability distribution followed by the random variable $X$? Specify its parameters.
2. Determine the probability that exactly 20 people among the 150 interviewed are allergic.
3. Determine the probability that at least $10\%$ of the people among the 150 interviewed are allergic.

In an effort to improve its sustainable development policy, a company conducted a statistical survey on its waste production.
In this survey, waste is classified into three categories:

• $69 \%$ of waste is mineral and non-hazardous;
• $28 \%$ of waste is non-mineral and non-hazardous;
• the remaining waste is hazardous waste.

This statistical survey also tells us that:

• $73 \%$ of mineral and non-hazardous waste is recyclable;
• $49 \%$ of non-mineral and non-hazardous waste is recyclable;
• $35 \%$ of hazardous waste is recyclable.

In this company, a piece of waste is randomly selected. We consider the following events:

• $M$ : ``The selected waste is mineral and non-hazardous'';
• N : ``The selected waste is non-mineral and non-hazardous'';
• $D$ : ``The selected waste is hazardous'';
• R: ``The selected waste is recyclable''.

We denote by $\bar{R}$ the complementary event of event $R$.
Part A

1. Copy and complete the probability tree below representing the situation described in the problem.
2. Justify that the probability that the waste is hazardous and recyclable is equal to 0.0105.
3. Determine the probability $P(M \cap \bar{R})$ and interpret the answer obtained in the context of the exercise.
4. Prove that the probability of event $R$ is $P(R) = 0.6514$.
5. Suppose that the selected waste is recyclable. Determine the probability that this waste is non-mineral and non-hazardous. Give the answer rounded to the ten-thousandth.

Part B
We recall that the probability that a randomly selected piece of waste is recyclable is equal to 0.6514.

1. In order to control the quality of collection in the company, a sample of 20 pieces of waste is randomly selected from production. We assume that the stock is sufficiently large to treat the sampling of this sample as drawing with replacement.
   We denote by $X$ the random variable equal to the number of recyclable pieces of waste in this sample. a. We admit that the random variable $X$ follows a binomial distribution. Specify its parameters. b. Give the probability that the sample contains exactly 14 recyclable pieces of waste. Give the answer rounded to the ten-thousandth.
2. In this question, we now select $n$ pieces of waste, where $n$ denotes a strictly positive natural number. a. Give the expression as a function of $n$ of the probability $p_n$ that no piece of waste in this sample is recyclable. b. Determine the value of the natural number $n$ from which the probability that at least one piece of waste in the sample is recyclable is greater than or equal to 0.9999.

A game offered at a fairground consists of making three successive shots at a moving target.
It has been observed that:

• If the player hits the target on one shot then they miss it on the next shot in $65\%$ of cases;
• If the player misses the target on one shot then they hit it on the next shot in $50\%$ of cases.

The probability that a player hits the target on their first shot is 0.6. For any event $A$, we denote $p(A)$ its probability and $\bar{A}$ the complementary event of $A$. We randomly choose a player for this shooting game. We consider the following events:

• $A_1$: ``The player hits the target on the $1^{\text{st}}$ shot''
• $A_2$: ``The player hits the target on the $2^{\mathrm{nd}}$ shot''
• $A_3$: ``The player hits the target on the $3^{\mathrm{rd}}$ shot''.

Part A

1. Copy and complete, with the corresponding probabilities on each branch, the probability tree below modelling the situation.

Let $X$ be the random variable that gives the number of times the player hits the target during the three shots.
2. Show that the probability that the player hits the target exactly twice during the three shots is equal to 0.4015.
3. The objective of this question is to calculate the expectation of the random variable $X$, denoted $E(X)$. a. Copy and complete the table below giving the probability distribution of the random variable $X$.

$X = x_i$	0	1	2	3
$p\left(X = x_i\right)$	0.1			0.0735

b. Calculate $E(X)$. c. Interpret the previous result in the context of the exercise.
Part B
We consider $N$, a natural number greater than or equal to 1.
A group of $N$ people comes to this stand to play this game under identical and independent conditions.
A player is declared a winner when they hit the target three times. We denote $Y$ the random variable that counts among the $N$ people the number of players declared winners.

1. In this question, $N = 15$. a. Justify that $Y$ follows a binomial distribution and determine its parameters. b. Give the probability, rounded to $10^{-3}$, that exactly 5 players win this game.
2. By the method of your choice, which you will explain, determine the minimum number of people who must come to this stand so that the probability that there is at least one winning player is greater than or equal to 0.98.

A boat rental company for tourism offers its clients two types of boats: sailboat and motorboat.
Furthermore, a client can take the PILOT option. In this case, the boat, whether sailboat or motorboat, is rented with a pilot.
We know that:

• $60\%$ of clients choose a sailboat; among them, $20\%$ take the PILOT option.
• $42\%$ of clients take the PILOT option.

A client is chosen at random and we consider the events:

• $V$: ``the client chooses a sailboat'';
• $L$: ``the client takes the PILOT option''.

Part A

1. Represent the situation with a probability tree that you will complete as you go.
2. Calculate the probability that the client chooses a sailboat and does not take the PILOT option.
3. Prove that the probability that the client chooses a motorboat and takes the PILOT option is equal to 0.30.
4. Deduce $P_{\bar{V}}(L)$, the probability of $L$ given that $V$ is not realized.
5. A client has taken the PILOT option. What is the probability that he chose a sailboat? Round to 0.01.

Part B
When a client does not take the PILOT option, the probability that his boat suffers a breakdown is equal to 0.12. This probability is only 0.005 if the client takes the PILOT option. We consider a client. We denote by $A$ the event: ``his boat suffers a breakdown''.

1. Determine $P(L \cap A)$ and $P(\bar{L} \cap A)$.
2. The company rents 1000 boats. How many breakdowns can it expect?

Part C
We recall that the probability that a given client takes the PILOT option is equal to 0.42. We consider a random sample of 40 clients. We denote by $X$ the random variable counting the number of clients in the sample taking the PILOT option.

1. We admit that the random variable $X$ follows a binomial distribution. Give its parameters without justification.
2. Calculate the probability, rounded to $10^{-3}$, that at least 15 clients take the PILOT option.