A store is equipped with self-service automatic checkouts where the customer scans their own items. The checkout software regularly triggers verification requests.
The check can be either ``complete'': the store employee then scans all of the customer's items again; or ``partial'': the employee then selects one or more of the customer's items to verify that they have been scanned correctly.
If a check is triggered, it is a complete check one time out of ten. When a complete check is triggered, a customer error is detected in $30\%$ of cases. When a partial check is performed, in $85\%$ of cases, there is no error.
A check is triggered at an automatic checkout. We consider the following events:

• T: ``The check is a complete check'';
• E: ``An error is detected during the check''.

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

1. Construct a probability tree representing the situation and then determine $P(\bar{T} \cap E)$.
2. Calculate the probability that an error is detected during a check.
3. Determine the probability that a complete check was performed, given that an error was detected. The answer will be given rounded to the nearest hundredth.

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: $$P \left( 7 < M _ { N } < 7{,}28 \right) \geqslant 0{,}95 .$$

In France there are two formulas for obtaining a driving license:

• Follow supervised driving training from age 15 for 2 years;
• Follow classical training (without supervised driving) from age 17.

In France currently, among young people who follow driving license training, $16\%$ choose supervised driving training, and among them, $74.7\%$ pass the driving test on their first attempt. Following classical training, the success rate on the first attempt is only $56.8\%$.
A young French person who has already taken the driving test is chosen at random, and we consider the following events $A$ and $R$:

• $A$: ``the young person followed supervised driving training'';
• $R$: ``the young person obtained their license on their first attempt''.

Results should be rounded to $10^{-3}$ if necessary.
Part A

1. Draw a probability tree modeling this situation.
2. a. Prove that $P(R) = 0.59664$.
   In the following, we will keep the value 0.597 rounded to $10^{-3}$. b. Give this result as a percentage and interpret it in the context of the exercise.
3. A young person who obtained their license on their first attempt is chosen. What is the probability that they followed supervised driving training?
4. What should be the proportion of young people following supervised driving training if we wanted the overall success rate (regardless of the training chosen) on the first attempt at the driving test to exceed $70\%$?

Part B
A driving school presents 10 candidates for the driving test for the first time, all of whom have followed supervised driving training. The fact of taking driving tests is modeled by independent random trials.
Let $X$ be the random variable giving the number of these 10 candidates who will obtain their license on their first attempt.

1. Justify that $X$ follows a binomial distribution with parameters $n = 10$ and $p = 0.747$.
2. Calculate $P(X \geqslant 6)$. Interpret this result.
3. Determine $E(X)$ and $V(X)$.
4. There are also 40 candidates who have not followed supervised driving training and who are taking the driving test for the first time. In the same way, let $Y$ be the random variable giving the number of these candidates who will obtain their license on their first attempt. We admit that $Y$ is independent of the variable $X$ and that in fact $E(Y) = 22.53$ and $V(Y) = 9.81$.
   Let $Z$ be the random variable counting the total number of candidates (among the 50) who will obtain their driving license on their first attempt at this driving school. a. Express $Z$ as a function of $X$ and $Y$. Deduce $E(Z)$ and $V(Z)$. b. Using Bienaymé-Chebyshev's inequality, show that the probability that there are fewer than 20 or more than 40 candidates who obtain their license on their first attempt is less than 0.12.

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 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.

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 company has a total of 60 employees, and each employee belongs to one of two departments, A or B. Department A has 20 employees and Department B has 40 employees. 50\% of the employees in Department A are women. 60\% of the women employees in the company belong to Department B. When one employee is randomly selected from the 60 employees and is found to belong to Department B, the probability that this employee is a woman is $p$. Find the value of $80p$. [4 points]

$49 \%$ of the packages contain clothing. Of the packages that are returns, $91 \%$ contain clothing.
A package is randomly selected. The following events are considered: $R$ : ``The package is a return.'' $K$ : ``The package contains clothing.'' Represent the described situation in a completely filled four-field table. For the case that the selected package is not a return, determine the probability that the package contains clothing.

An air quality measurement station measures levels of $\mathrm { NO } _ { 2 }$ and suspended particles. The probability that on a given day a level of $\mathrm { NO } _ { 2 }$ exceeding the permitted level is measured is 0.16. On days when the permitted level of $\mathrm { NO } _ { 2 }$ is exceeded, the probability that the permitted level of particles is exceeded is 0.33. On days when the level of $\mathrm { NO } _ { 2 }$ is not exceeded, the probability that the level of particles is exceeded is 0.08.\ a) ( 0.5 points) What is the probability that on a given day both permitted levels are exceeded?\ b) ( 0.75 points) What is the probability that at least one of the two is exceeded?\ c) ( 0.5 points) Are the events ``on a given day the permitted level of $\mathrm { NO } _ { 2 }$ is exceeded'' and ``on a given day the permitted level of particles is exceeded'' independent?\ d) ( 0.75 points) What is the probability that on a given day the permitted level of $\mathrm { NO } _ { 2 }$ is exceeded, given that the permitted level of particles has not been exceeded?

A company markets three types of products A, B and C. Four out of every seven products are of type A, two out of every seven products are of type B and the rest are of type C. For export, 40\% of type A products, 60\% of type B products and 20\% of type C products are destined. A product is chosen at random, it is requested: a) (1.25 points) Calculate the probability that the product is destined for export. b) (1.25 points) Calculate the probability that it is of type C given that the product is destined for export.

According to a certain country's investigation of missing light aircraft: 70\% of missing light aircraft are eventually found. Among the aircraft that are found, 60\% have emergency locator transmitters installed; among the missing aircraft that are not found, 90\% do not have emergency locator transmitters installed. Emergency locator transmitters send signals when the aircraft crashes, allowing rescue personnel to locate it. A light aircraft is now missing. If it is known that the aircraft has an emergency locator transmitter installed, the probability that it will be found is (15--1)(15--2).

All senior high school students at a certain school have taken either Mathematics A or Mathematics B on the scholastic aptitude test. Among these students, those taking only Mathematics A account for $\frac { 3 } { 10 }$ of all senior high school students. Among students taking Mathematics A, $\frac { 5 } { 8 }$ also took Mathematics B. What is the proportion of students taking only Mathematics B among all students at the school taking Mathematics B? (Express as a fraction in lowest terms)

Problem 6

A product factory manufactures 2 types of products: product-I and product-II. Part-A is necessary for product-I, and both part-$A$ and part-$B$ are necessary for product-II. There are parts that have standard quality and parts that do not have standard quality among part-$A$ and part-$B$. All parts are delivered from the part factory to the product factory, but there is no quality check of any part. The qualities of part-$A$ and part-$B$ are independent, and they will not affect each other. The probabilities that part-$A$ and part-$B$ have standard quality are $a$ and $b$, respectively.
A final quality inspection is made in the product factory for product-I and for product-II before shipment. The inspection judges whether the quality of each product meets the standard or not. The inspections will not affect each other. The product inspection is not perfect: namely, products that have standard quality pass the product inspection as acceptable with the probability $x$. The products that do not have standard quality pass the product inspection as acceptable with the probability $y$.
Answer the following questions: I. A product-I is randomly sampled and inspected once. Here, the probability that product-I can be manufactured with standard quality is defined as follows:

• The probability that product-I has standard quality is $c$ if part-$A$ has standard quality.
• Product-I will never have standard quality if part-A does not have standard quality.

1. Show the probability that the selected product-I passes the product inspection as acceptable.
2. Show the probability that the selected product-I actually has standard quality after it has passed the product inspection as acceptable.

II. A product-II is randomly sampled and inspected $n$ times. Here, the probability that product-II can be manufactured with standard quality is defined as follows:

• The probability that product-II has standard quality is $c$ if both part-$A$ and part-$B$ have standard quality.
• The probability that product-II has standard quality is $d$ if only either part-$A$ or part-$B$ has standard quality.
• Product-II will never have standard quality if both part-$A$ and part-$B$ do not have standard quality.

1. Show the probability that the selected product-II has standard quality.
2. Show the probability that the selected product-II actually has standard quality after it has passed all product inspections (i.e., $n$ times) as acceptable.