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. $N _ { 0 } = 20$

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.

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.

Alice has two urns A and B each containing four indistinguishable balls. Urn A contains two green balls and two red balls. Urn B contains three green balls and one red ball. Alice randomly chooses an urn and then a ball from that urn. She obtains a green ball. The probability that she chose urn B is:
A. $\frac{3}{8}$
B. $\frac{1}{2}$
C. $\frac{3}{5}$
D. $\frac{5}{8}$

A video game rewards players who have won a challenge with a randomly drawn object. The drawn object can be ``common'' or ``rare''. Two types of objects, common or rare, are available: swords and shields.
The video game designers have planned that:

• the probability of drawing a rare object is $7\%$;
• if a rare object is drawn, the probability that it is a sword is $80\%$;
• if a common object is drawn, the probability that it is a sword is $40\%$.

Part A
A player has just won a challenge and draws an object at random. We denote:

• R the event ``the player draws a rare object'';
• $E$ the event ``the player draws a sword'';
• $\bar{R}$ and $\bar{E}$ the complementary events of events $R$ and $E$.

1. Draw a probability tree modelling the situation, then calculate $P(R \cap E)$.
2. Calculate the probability of drawing a sword.
3. The player has drawn a sword. Determine the probability that it is a rare object. Round the result to the nearest thousandth.

A video game rewards players who have won a challenge with a randomly drawn object. The drawn object can be ``common'' or ``rare''. Two types of objects, common or rare, are available: swords and shields.
The video game designers have planned that:

• the probability of drawing a rare object is $7\%$;
• if a rare object is drawn, the probability that it is a sword is $80\%$;
• if a common object is drawn, the probability that it is a sword is $40\%$.

Part B
A player wins 30 challenges. We denote $X$ the random variable corresponding to the number of rare objects the player obtains after winning 30 challenges. The successive draws are considered independent.

1. Determine, by justifying, the probability distribution followed by the random variable $X$. Specify its parameters, as well as its expected value.
2. Determine $P(X < 6)$. Round the result to the nearest thousandth.
3. Determine the largest value of $k$ such that $P(X \geqslant k) \geqslant 0.5$. Interpret the result in the context of the exercise.
4. The video game developers want to offer players the option to buy a ``gold ticket'' which allows them to draw $N$ objects. The probability of drawing a rare object remains $7\%$. The developers would like that by buying a gold ticket, the probability that a player obtains at least one rare object in these $N$ draws is greater than or equal to $0.95$. Determine the minimum number of objects to draw to achieve this objective. Care should be taken to detail the approach used.

Data published on March 1st, 2023 by the Ministry of Ecological Transition on the registration of private vehicles in France in 2022 contain the following information:

• $22.86\%$ of vehicles were new vehicles;
• $8.08\%$ of new vehicles were rechargeable hybrids;
• $1.27\%$ of used vehicles (that is, those that are not new) were rechargeable hybrids.

Throughout the exercise, probabilities will be rounded to the ten-thousandth.
Part I
In this part, we consider a private vehicle registered in France in 2022. We denote:

• $N$ the event ``the vehicle is new'';
• $R$ the event ``the vehicle is a rechargeable hybrid'';
• $\bar{N}$ and $\bar{R}$ the complementary events of $N$ and $R$.

1. Represent the situation with a probability tree.
2. Calculate the probability that this vehicle is new and a rechargeable hybrid.
3. Prove that the value rounded to the ten-thousandth of the probability that this vehicle is a rechargeable hybrid is 0.0283.
4. Calculate the probability that this vehicle is new given that it is a rechargeable hybrid.

Part II
In this part, we choose 500 private rechargeable hybrid vehicles registered in France in 2022. In what follows, we will assume that the probability that such a vehicle is new is equal to 0.65. We treat the choice of these 500 vehicles as a random draw with replacement. We call $X$ the random variable representing the number of new vehicles among the 500 vehicles chosen.

1. We assume that the random variable $X$ follows a binomial distribution. Specify the values of its parameters.
2. Determine the probability that exactly 325 of these vehicles are new.
3. Determine the probability $p(X \geq 325)$ then interpret the result in the context of the exercise.

Part III
We now choose $n$ private rechargeable hybrid vehicles registered in France in 2022, where $n$ denotes a strictly positive natural number. We recall that the probability that such a vehicle is new is equal to 0.65. We treat the choice of these $n$ vehicles as a random draw with replacement.

1. Give the expression as a function of $n$ of the probability $p_n$ that all these vehicles are used.
2. We denote $q_n$ the probability that at least one of these vehicles is new. By solving an inequality, determine the smallest value of $n$ such that $q_n \geqslant 0.9999$.

A survey conducted in France provides the following information:

• $60\%$ of people over 15 years old intend to watch the Paris 2024 Olympic and Paralympic Games (OPG) on television;
• among those who intend to watch the OPG, 8 out of 9 people declare that they regularly practice a sport.

A person over 15 years old is chosen at random. The following events are considered:

• $J$: ``the person intends to watch the Paris 2024 OPG on television'';
• $S$: ``the chosen person declares that they regularly practice a sport''.

We denote by $\bar{J}$ and $\bar{S}$ their complementary events.
In questions 1. and 2., probabilities will be given in the form of an irreducible fraction.

1. Demonstrate that the probability that the chosen person intends to watch the Paris 2024 OPG on television and declares that they regularly practice a sport is $\frac{8}{15}$. A weighted tree diagram may be used.
   According to this survey, two out of three people over 15 years old declare that they regularly practice a sport.
   
2. [2.] a. Calculate the probability that the chosen person does not intend to watch the Paris 2024 OPG on television and declares that they regularly practice a sport. b. Deduce the probability of $S$ given $\bar{J}$ denoted $P_{\bar{J}}(S)$.
   In the rest of the exercise, results will be rounded to the nearest thousandth.
   
3. [3.] As part of a promotional operation, 30 people over 15 years old are chosen at random. This choice is treated as sampling with replacement. Let $X$ be the random variable that gives the number of people declaring that they regularly practice a sport among the 30 people. a. Determine the nature and parameters of the probability distribution followed by $X$. b. Calculate the probability that exactly 16 people declare that they regularly practice a sport among the 30 people. c. The French judo federation wishes to offer a ticket for the final of the mixed team judo event at the Arena Champ-de-Mars for each person declaring that they regularly practice a sport among these 30 people. The price of a ticket is $380\,€$ and a budget of 10000 euros is available for this operation. What is the probability that this budget is insufficient?

A marketing agency studied customer satisfaction regarding customer service when purchasing a television. These purchases were made either online, in an appliance store chain, or in a large supermarket. Online purchases represent $60 \%$ of sales, appliance store purchases $30 \%$ of sales, and large supermarket purchases $10 \%$ of sales. A survey shows that the proportion of customers satisfied with customer service is:

• $75 \%$ for online customers;
• $90 \%$ for appliance store customers;
• $80 \%$ for large supermarket customers.

A customer who purchased the television model in question is chosen at random. The following events are defined:

• I: ``the customer made their purchase online'';
• $M$: ``the customer made their purchase in an appliance store'';
• $G$: ``the customer made their purchase in a large supermarket'';
• S: ``the customer is satisfied with customer service''.

If $A$ is any event, we denote by $\bar { A }$ its complementary event and $P ( A )$ its probability.

1. Reproduce and complete the tree diagram opposite.
2. Calculate the probability that the customer made their purchase online and is satisfied with customer service.
3. Prove that $P ( S ) = 0.8$.
4. A customer is satisfied with customer service. What is the probability that they made their purchase online? Give the result rounded to $10 ^ { - 3 }$.
5. To conduct the study, the agency must contact 30 customers each day among the television buyers. We assume that the number of customers is large enough to treat the choice of 30 customers as sampling with replacement. Let $X$ be the random variable that, for each sample of 30 customers, associates the number of customers satisfied with customer service. a. Justify that $X$ follows a binomial distribution and specify its parameters. b. Determine the probability, rounded to $10 ^ { - 3 }$, that at least 25 customers are satisfied in a sample of 30 customers contacted on the same day.
6. By solving an inequality, determine the minimum sample size of customers to contact so that the probability that at least one of them is not satisfied is greater than $0.99$.
7. In the two questions a. and b. that follow, we are only interested in online purchases. When a television order is placed by a customer, the delivery time of the television is modeled by a random variable $T$ equal to the sum of two random variables $T _ { 1 }$ and $T _ { 2 }$.

The random variable $T _ { 1 }$ models the integer number of days for the television to be transported from a storage warehouse to a distribution platform. The random variable $T _ { 2 }$ models the integer number of days for the television to be transported from this platform to the customer's home.
We admit that the random variables $T _ { 1 }$ and $T _ { 2 }$ are independent, and we are given:

• The expectation $E \left( T _ { 1 } \right) = 4$ and the variance $V \left( T _ { 1 } \right) = 2$;
• The expectation $E \left( T _ { 2 } \right) = 3$ and the variance $V \left( T _ { 2 } \right) = 1$. a. Determine the expectation $E ( T )$ and the variance $V ( T )$ of the random variable $T$. b. A customer places a television order online. Justify that the probability that they receive their television between 5 and 9 days after their order is greater than or equal to $\frac { 2 } { 3 }$.

In the journal Lancet Public Health, researchers claim that on May 11, 2020, 5.7\% of French adults had already been infected with COVID 19.
A test has been implemented: this allows to determine (even long after infection), whether or not a person has already been infected with COVID 19. If the test is positive, this means that the person has already been infected with COVID 19.
The sensitivity of a test is the probability that it is positive given that the person has been infected with the disease. The specificity of a test is the probability that the test is negative given that the person has not been infected with the disease.
The test manufacturer provides the following characteristics:

• Its sensitivity is 0.8.
• Its specificity is 0.99.

An individual is drawn and subjected to the test from the adult French population on May 11, 2020. Let $T$ be the event ``the test performed is positive''.

1. Complete the probability tree with the data from the statement.
2. Show that $p(T) = 0.05503$.
3. What is the probability that an individual has been infected given that their test is positive? Give an approximate value to $10^{-4}$ near of the result.

We consider a group from the population of another country subjected to the same test with sensitivity 0.8 and specificity 0.99.
In this group the proportion of individuals with a positive test is 29.44\%.
An individual is chosen at random from this group; what is the probability that they have been infected?

A student eats every day at the university restaurant. This restaurant offers vegetarian and non-vegetarian dishes.

• When on a given day the student has chosen a vegetarian dish, the probability that he chooses a vegetarian dish the next day is 0.9.
• When on a given day the student has chosen a non-vegetarian dish, the probability that he chooses a vegetarian dish the next day is 0.7.

For any natural number $n$, we denote by $V _ { n }$ the event ``the student chose a vegetarian dish on the $n ^ { \mathrm { th } }$ day'' and $p _ { n }$ the probability of $V _ { n }$. On the first day of the semester, the student chose the vegetarian dish. Thus $p _ { 1 } = 1$.

1. a. Indicate the value of $p _ { 2 }$. b. Show that $p _ { 3 } = 0.88$. You may use a probability tree. c. Given that on the 3rd day the student chose a vegetarian dish, what is the probability that he chose a non-vegetarian dish the previous day? Round the result to $10 ^ { - 2 }$.
2. Copy and complete the probability tree.
3. Justify that, for any natural number $n \geqslant 1 , p _ { n + 1 } = 0.2 p _ { n } + 0.7$.
4. We wish to have the list of the first terms of the sequence $( p _ { n } )$ for $n \geqslant 1$. For this, we use a function called meals programmed in Python language, of which three versions are proposed below.

\begin{verbatim} Program 1 def meals(n): p=1 L= [p] for k in range(1,n): p = 0.2*p+0.7 L. append(p) return(L) \end{verbatim}
\begin{verbatim} Program 2 def meals(n): p=1 L= [p] for k in range(1,n+1): p = 0.2*p+0.7 L. append(p) return(L) \end{verbatim}
\begin{verbatim} Program 3 def meals(n): p=1 L=[p] for k in range(1,n): p = 0.2*p+0.7 L.append(p+1) return(L) \end{verbatim}
a. Which of these programs allows displaying the first $n$ terms of the sequence $\left( p _ { n } \right)$? No justification is required. b. With the program chosen in question a., give the result displayed for $n = 5$.
4. Prove by induction that, for any natural number $n \geqslant 1 , p _ { n } = 0.125 \times 0.2 ^ { n - 1 } + 0.875$.
5. Deduce the limit of the sequence $\left( p _ { n } \right)$.

In tennis, the player who is serving can, in case of failure on the first serve, serve a second ball. In match play, Abel succeeds with his first serve in $70\%$ of cases. When the first serve is successful, he wins the point in $80\%$ of cases. On the other hand, after a failure on his first serve, Abel wins the point in $45\%$ of cases. Abel is serving. Consider the following events:

• S: ``Abel succeeds with his first serve''
• G: ``Abel wins the point''.

1. Describe the event $S$ then translate the situation with a probability tree.
2. Calculate $P(S \cap G)$.
3. Justify that the probability of event $G$ is equal to 0.695.
4. Abel has won the point. What is the probability that he succeeded with his first serve?
5. Are events $S$ and $G$ independent? Justify.

There are four blood groups in the human species: $\mathrm { A } , \mathrm { B } , \mathrm { AB }$ and O. Each blood group can present a rhesus factor. When it is present, we say that the rhesus is positive, otherwise we say that it is negative.
Within the French population, we know that:

• $45 \%$ of individuals belong to group A, and among them $85 \%$ are rhesus positive;
• $10 \%$ of individuals belong to group B, and among them $84 \%$ are rhesus positive;
• $3 \%$ of individuals belong to group AB, and among them $82 \%$ are rhesus positive.

We randomly choose a person from the French population. We denote by:

• A the event ``The chosen person is of blood group A'';
• B the event ``The chosen person is of blood group B'';
• $AB$ the event ``The chosen person is of blood group AB'';
• O the event ``The chosen person is of blood group O'';
• $R$ the event ``The chosen person has a positive rhesus factor''.

For any event $E$, we denote by $\bar { E }$ the complementary event of $E$ and $p ( E )$ the probability of $E$.

1. Copy the tree opposite and complete the ten blanks.
2. Show that $p ( B \cap R ) = 0{,}084$. Interpret this result in the context of the exercise.
3. We specify that $p ( R ) = 0{,}8397$. Show that $p _ { O } ( R ) = 0{,}83$.
4. We say that an individual is a ``universal donor'' when their blood can be transfused to any person without risk of incompatibility. Blood group O with negative rhesus is the only one satisfying this characteristic. Show that the probability that an individual randomly chosen from the French population is a universal donor is 0,0714.
5. During a blood collection, a sample of 100 people is chosen from the population of a French city. This population is large enough to assimilate this choice to sampling with replacement. We denote by $X$ the random variable that associates to each sample of 100 people the number of universal donors in that sample. a. Justify that $X$ follows a binomial distribution and specify its parameters. b. Determine to $10 ^ { - 3 }$ near the probability that there are at most 7 universal donors in this sample. c. Show that the expectation $E ( X )$ of the random variable $X$ is equal to 7,14 and that its variance $V ( X )$ is equal to 6,63 to $10 ^ { - 2 }$ near.
6. During the national blood donation week, a blood collection is organized in $N$ randomly chosen French cities numbered $1,2,3 , \ldots , N$ where $N$ is a non-zero natural integer. We consider the random variable $X _ { 1 }$ which associates to each sample of 100 people from city 1 the number of universal donors in that sample. We define in the same way the random variables $X _ { 2 }$ for city $2 , \ldots , X _ { N }$ for city $N$. We assume that these random variables are independent and that they have the same expectation equal to 7,14 and the same variance equal to 6,63. We consider the random variable $M _ { N } = \frac { X _ { 1 } + X _ { 2 } + \ldots + X _ { N } } { N }$. a. What does the random variable $M _ { N }$ represent in the context of the exercise? b. Calculate the expectation $E \left( M _ { N } \right)$. c. We denote by $V \left( M _ { N } \right)$ the variance of the random variable $M _ { N }$. Show that $V \left( M _ { N } \right) = \frac { 6{,}63 } { N }$. d. Determine the smallest value of $N$ for which the Bienaymé-Chebyshev inequality allows us to assert that: $$P \left( 7 < M _ { N } < 7{,}28 \right) \geqslant 0{,}95 .$$

An American team mapped food allergies in children in the United States for the first time in 2020. It is known that in 2020, $17\%$ of the population of the United States lives in rural areas and $83\%$ in urban areas. Among children in the United States living in rural areas, $6.2\%$ are affected by food allergies. Also, $9\%$ of children in the United States are affected by food allergies.
For any event $E$, we denote $P(E)$ its probability and $\bar{E}$ its complementary event. Unless otherwise stated, probabilities will be given in exact form.
A child is randomly selected from the population of the United States and we denote:

• R The event: ``the child interviewed lives in a rural area'';
• A The event: ``the child interviewed is affected by food allergies''.

Part A

1. Translate this situation using a probability tree. This tree may be completed later.
2. a. Calculate the probability that the child interviewed lives in a rural area and is affected by food allergies. b. Deduce the probability that the child interviewed lives in an urban area and is affected by food allergies. c. The child interviewed lives in an urban area. What is the probability that he/she is affected by food allergies? Round the result to $10^{-4}$.

Part B
A study is conducted by randomly interviewing 100 children in the United States. We assume that this choice amounts to successive independent draws with replacement. We denote $X$ the random variable giving the number of children affected by food allergies in the sample considered.

1. Justify that the random variable $X$ follows a binomial distribution and specify its parameters.
2. What is the probability that at least 10 children among the 100 interviewed are affected by food allergies? Round the result to $10^{-4}$.

Part C
We are interested in a sample of 20 children affected by food allergies chosen at random. The age of onset of the first allergic symptoms of these 20 children is modeled by the random variables $A_1, A_2, \ldots, A_{20}$. We assume that these random variables are independent and follow the same distribution with expectation 4 and variance 2.25. We consider the random variable: $$M_{20} = \frac{A_1 + A_2 + \ldots + A_{20}}{20}.$$

1. What does the random variable $M_{20}$ represent in the context of the exercise?
2. Determine the expectation and variance of $M_{20}$.
3. Justify, using the concentration inequality, that $$P\left(2 < M_{20} < 6\right) > 0.97.$$ Interpret this result in the context of the exercise.

A company that manufactures toys must perform conformity checks before their commercialization. In this exercise, we are interested in two tests performed by the toy company: a manufacturing test and a safety test. Following a large number of verifications, the company claims that:

• $95 \%$ of toys pass the manufacturing test;
• Among toys that pass the manufacturing test, $98 \%$ pass the safety test;
• $1 \%$ of toys pass neither of the two tests.

A toy is chosen at random from the toys produced. We denote:

• F the event: ``the toy passes the manufacturing test'';
• S the event: ``the toy passes the safety test''.

Part A

1. From the data in the statement, give the probabilities $P ( F )$ and $P _ { F } ( S )$.
2. a. Construct a probability tree that illustrates the situation with the data available in the statement. b. Show that $P _ { \bar { F } } ( \bar { S } ) = 0.2$.
3. Calculate the probability that the chosen toy passes both tests.
4. Show that the probability that the toy passes the safety test is 0.97 rounded to the nearest hundredth.
5. When the toy has passed the safety test, what is the probability that it passes the manufacturing test? Give an approximate value of the result to the nearest hundredth.

Part B
A batch of $n$ toys is randomly selected from the company's production, where $n$ is a strictly positive integer. We assume that this selection is made from a sufficiently large quantity of toys to be assimilated to a succession of $n$ independent draws with replacement. Recall that the probability that a toy passes the manufacturing test is equal to 0.95. Let $S _ { n }$ be the random variable that counts the number of toys that have passed the manufacturing test. We admit that $S _ { n }$ follows the binomial distribution with parameters $n$ and $p = 0.95$.

1. Express the expectation and variance of the random variable $S _ { n }$ as a function of $n$.
2. In this question, we set $n = 150$. a. Determine an approximate value to $10 ^ { - 3 }$ of $P \left( S _ { 150 } = 145 \right)$. Interpret this result in the context of the exercise. b. Determine the probability that at least $94 \%$ of the toys in this batch pass the manufacturing test. Give an approximate value of the result to $10 ^ { - 3 }$.
3. In this question, the non-zero natural integer $n$ is no longer fixed.

Let $F _ { n }$ be the random variable defined by: $F _ { n } = \frac { S _ { n } } { n }$. The random variable $F _ { n }$ represents the proportion of toys that pass the manufacturing test in a batch of $n$ toys selected. We denote $E \left( F _ { n } \right)$ the expectation and $V \left( F _ { n } \right)$ the variance of the random variable $F _ { n }$. a. Show that $E \left( F _ { n } \right) = 0.95$ and that $V \left( F _ { n } \right) = \frac { 0.0475 } { n }$. b. We are interested in the following event $I$: ``the proportion of toys that pass the manufacturing test in a batch of $n$ toys is strictly between $93 \%$ and $97 \%$''. Using the Bienaymé-Chebyshev inequality, determine a value $n$ of the size of the batch of toys to select, from which the probability of event $I$ is greater than or equal to 0.96.

A company that manufactures toys must perform conformity checks before commercialization. In this exercise, we are interested in two tests performed by the company: a so-called manufacturing test and a so-called safety test.
Following a large number of verifications, the company claims that:

• $95 \%$ of toys pass the manufacturing test;
• among toys that pass the manufacturing test, $98 \%$ pass the safety test;
• $1 \%$ of toys pass neither of the two tests.

A toy is chosen at random from the toys produced. We denote:

• $F$ the event: ``the toy passes the manufacturing test'';
• $S$ the event: ``the toy passes the safety test''.

Part A

1. From the data in the statement, give the probabilities $P ( F )$ and $P _ { F } ( S )$.
2. (a) Construct a probability tree that illustrates the situation with the data available in the statement.
   (b) Show that $P _ { \bar { F } } ( \bar { S } ) = 0.2$.
3. Calculate the probability that the chosen toy passes both tests.
4. Show that the probability that the toy passes the safety test is 0.97 rounded to the nearest hundredth.
5. When the toy has passed the safety test, what is the probability that it passes the manufacturing test? Give an approximate value of the result to the nearest hundredth.

Part B

A batch of $n$ toys is randomly selected from the company's production, where $n$ is a strictly positive integer. We assume that this selection is made from a sufficiently large quantity of toys to be assimilated to a succession of $n$ independent draws with replacement.
We recall that the probability that a toy passes the manufacturing test is equal to 0.95. Let $S _ { n }$ be the random variable that counts the number of toys that have passed the manufacturing test. We admit that $S _ { n }$ follows the binomial distribution with parameters $n$ and $p = 0.95$.

1. Express the expectation and variance of the random variable $S _ { n }$ as a function of $n$.
2. In this question, we set $n = 150$.
   (a) Determine an approximate value to $10 ^ { - 3 }$ of $P \left( S _ { 150 } = 145 \right)$. Interpret this result in the context of the exercise.
   (b) Determine the probability that at least $94 \%$ of the toys in this batch pass the manufacturing test. Give an approximate value of the result to $10 ^ { - 3 }$.
3. In this question, the non-zero natural number $n$ is no longer fixed.

Let $F _ { n }$ be the random variable defined by: $F _ { n } = \frac { S _ { n } } { n }$. The random variable $F _ { n }$ represents the proportion of toys that pass the manufacturing test in a batch of $n$ toys selected. We denote $E \left( F _ { n } \right)$ the expectation and $V \left( F _ { n } \right)$ the variance of the random variable $F _ { n }$.
(a) Show that $E \left( F _ { n } \right) = 0.95$ and that $V \left( F _ { n } \right) = \frac { 0.0475 } { n }$.
(b) We are interested in the following event $I$: ``the proportion of toys that pass the manufacturing test in a batch of $n$ toys is strictly between $93 \%$ and $97 \%$''. Using the Bienaymé-Chebyshev inequality, determine a value $n$ of the size of the batch of toys to be selected, from which the probability of event $I$ is greater than or equal to 0.96.

In a school with 1200 students, a survey was conducted on their knowledge of two foreign languages, English and Spanish.
In this survey, it was found that 600 students speak English, 500 speak Spanish, and 300 do not speak either of these languages.
If a student from this school is chosen at random and it is known that he does not speak English, what is the probability that this student speaks Spanish?
(A) $\frac{1}{2}$ (B) $\frac{5}{8}$ (C) $\frac{1}{4}$ (D) $\frac{5}{6}$ (E) $\frac{5}{14}$

A resident of a metropolitan region has a 50\% probability of being late for work when it rains in the region; if it does not rain, his probability of being late is 25\%. For a given day, the meteorological service estimates a 30\% probability of rain occurring in that region.
What is the probability that this resident will be late for work on the day for which the rain estimate was given?
(A) 0.075
(B) 0.150
(C) 0.325
(D) 0.600
(E) 0.800

A test developed to detect Covid gives the correct diagnosis for $99\%$ of people with Covid. It also gives the correct diagnosis for $99\%$ of people without Covid. In a city $\frac{1}{1000}$ of the population has Covid.
If the probability is $x\%$, then your answer should be the integer closest to $x$. E.g., for probability $\frac{1}{3} = 33.33\ldots\%$, you should type 33 as your answer. For probability $\frac{2}{3}$ you should type 67 as your answer.
What is the probability that a randomly selected person tests positive? (We assume that in our random selection every person is equally likely to be chosen.) [2 points]

A test developed to detect Covid gives the correct diagnosis for $99\%$ of people with Covid. It also gives the correct diagnosis for $99\%$ of people without Covid. In a city $\frac{1}{1000}$ of the population has Covid.
If the probability is $x\%$, then your answer should be the integer closest to $x$. E.g., for probability $\frac{1}{3} = 33.33\ldots\%$, you should type 33 as your answer. For probability $\frac{2}{3}$ you should type 67 as your answer.
Suppose that a randomly selected person tested positive. What is the probability that this person has Covid? [2 points]

A discrete random variable $X$ can take values $0,1,2,3,4,5,6,7$ and its probability mass function is $$\mathrm { P } ( X = x ) = \left\{ \begin{array} { l l } c , & x = 0,1,2 \\ 2 c , & x = 3,4,5 \\ 5 c ^ { 2 } , & x = 6,7 \end{array} \quad ( \text { where } c \text { is a positive number } ) \right.$$ Let $A$ be the event that the random variable $X$ is at least 6, and let $B$ be the event that the random variable $X$ is at least 3. What is the value of $\mathrm { P } ( A \mid B )$? [3 points]
(1) $\frac { 1 } { 5 }$
(2) $\frac { 1 } { 6 }$
(3) $\frac { 1 } { 7 }$
(4) $\frac { 1 } { 8 }$
(5) $\frac { 1 } { 9 }$

A certain class consists of 18 male students and 16 female students. All students in this class take a class in either Chinese or Japanese, but not both. Among the male students, 12 take Chinese class, and among the female students, 7 take Japanese class. When a student selected from this class is taking Chinese class, what is the probability that this student is female? [3 points]
(1) $\frac { 1 } { 7 }$
(2) $\frac { 2 } { 7 }$
(3) $\frac { 3 } { 7 }$
(4) $\frac { 4 } { 7 }$
(5) $\frac { 5 } { 7 }$

For two events $A , B$, $$\mathrm { P } ( A ) = \frac { 1 } { 4 } , \quad \mathrm { P } ( B ) = \frac { 2 } { 3 } , \quad A \subset B$$ What is the value of $\mathrm { P } ( A \mid B )$? [3 points]
(1) $\frac { 1 } { 8 }$
(2) $\frac { 1 } { 4 }$
(3) $\frac { 3 } { 8 }$
(4) $\frac { 1 } { 2 }$
(5) $\frac { 5 } { 8 }$