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 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.

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

There are four distinct balls labelled $1, 2, 3, 4$ and four distinct bins labelled A, B, C, D. The balls are picked up in order and placed into one of the four bins at random. Let $E_i$ denote the event that the first $i$ balls go into distinct bins. Calculate the following probabilities.
(i) $\Pr[E_4]$
(ii) $\Pr[E_4 \mid E_3]$
(iii) $\Pr[E_4 \mid E_2]$
(iv) $\Pr[E_3 \mid E_4]$.
Notation: $\Pr[X] =$ the probability of event $X$ taking place. $\Pr[X \mid Y] =$ the probability of event $X$ taking place, given that event $Y$ has taken place.

Out of the 14 students taking a test, 5 are well prepared, 6 are adequately prepared and 3 are poorly prepared. There are 10 questions on the test paper. A well prepared student can answer 9 questions correctly, an adequately prepared student can answer 6 questions correctly and a poorly prepared student can answer only 3 questions correctly.
For each probability below, write your final answer as a rational number in lowest form.
(a) If a randomly chosen student is asked two distinct randomly chosen questions from the test, what is the probability that the student will answer both questions correctly?
Note: The student and the questions are chosen independently of each other. "Random" means that each individual student/each pair of questions is equally likely to be chosen.
(b) Now suppose that a student was chosen at random and asked two randomly chosen questions from the exam, and moreover did answer both questions correctly. Find the probability that the chosen student was well prepared.

Solve the following two independent problems.
(i) A mother and her two daughters participate in a game show. At first, the mother tosses a fair coin.
Case 1: If the result is heads, then all three win individual prizes and the game ends. Case 2: If the result is tails, then each daughter separately throws a fair die and wins a prize if the result of her die is 5 or 6. (Note that in case 2 there are two independent throws involved and whether each daughter gets a prize or not is unaffected by the other daughter's throw.)
(a) Suppose the first daughter did not win a prize. What is the probability that the second daughter also did not win a prize?
(b) Suppose the first daughter won a prize. What is the probability that the second daughter also won a prize?
(ii) Prove or disprove each of the following statements.
(a) $2 ^ { 40 } > 20!$
(b) $1 - \frac { 1 } { x } \leq \ln x \leq x - 1$ for all $x > 0$.

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 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) $\frac { 5 } { 8 }$

Bag A contains 5 cards with the numbers $1,2,3,4,5$ written on them, one each, and Bag B contains 5 cards with the numbers $6,7,8,9,10$ written on them, one each. One card is randomly drawn from each of the two bags A and B. When the sum of the two numbers on the drawn cards is odd, what is the probability that the number on the card drawn from Bag A is even? [3 points]
(1) $\frac { 5 } { 13 }$
(2) $\frac { 4 } { 13 }$
(3) $\frac { 3 } { 13 }$
(4) $\frac { 2 } { 13 }$
(5) $\frac { 1 } { 13 }$

For two events $A$ and $B$, $\mathrm { P } ( A ) = \frac { 1 } { 2 } , \mathrm { P } \left( B ^ { C } \right) = \frac { 2 } { 3 }$, and $\mathrm { P } ( B \mid A ) = \frac { 1 } { 6 }$. What is the value of $\mathrm { P } \left( A ^ { C } \mid B \right)$? (Here, $A ^ { C }$ is the complement of $A$.) [3 points]
(1) $\frac { 1 } { 2 }$
(2) $\frac { 7 } { 12 }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 3 } { 4 }$
(5) $\frac { 5 } { 6 }$

10\% of the emails Chulsu received contain the word ``travel''. 50\% of emails containing ``travel'' are advertisements, and 20\% of emails not containing ``travel'' are advertisements. Given that an email Chulsu received is an advertisement, what is the probability that it contains the word ``travel''? [3 points]
(1) $\frac { 5 } { 23 }$
(2) $\frac { 6 } { 23 }$
(3) $\frac { 7 } { 23 }$
(4) $\frac { 8 } { 23 }$
(5) $\frac { 9 } { 23 }$

10\% of the emails Cheol-su receives contain the word ``travel.'' 50\% of emails containing ``travel'' are advertisements, and 20\% of emails not containing ``travel'' are advertisements. Given that an email Cheol-su received is an advertisement, what is the probability that this email contains the word ``travel''? [3 points]
(1) $\frac { 5 } { 23 }$
(2) $\frac { 6 } { 23 }$
(3) $\frac { 7 } { 23 }$
(4) $\frac { 8 } { 23 }$
(5) $\frac { 9 } { 23 }$

[Probability and Statistics] A training facility has three courses A, B, and C to be experienced in order, with the entrance and exit being the same. There are 30 envelopes at course A, 60 envelopes at course B, and 90 envelopes at course C. Each envelope contains 1, 2, or 3 coupons. The following table shows the number of envelopes by the number of coupons for each course.

\multicolumn{1}{|c|}{Number of Coupons}	1	2	3	Total
A	20	10	0	30
B	30	20	10	60
C	40	30	20	90

After completing each course, a student randomly selects one envelope from that course and receives the coupons inside. A student who started first completed all three courses and received a total of 4 coupons. What is the probability that the student received 2 coupons at course B? [3 points]
(1) $\frac { 6 } { 23 }$
(2) $\frac { 8 } { 23 }$
(3) $\frac { 10 } { 23 }$
(4) $\frac { 12 } { 23 }$
(5) $\frac { 14 } { 23 }$

Bag A contains 5 cards with the numbers $1,2,3,4,5$ written on them, and Bag B contains 6 cards with the numbers $1,2,3,4,5,6$ written on them. A die is rolled once. If the result is a multiple of 3, a card is randomly drawn from Bag A; otherwise, a card is randomly drawn from Bag B. Given that the number on the card drawn from the bag is even, what is the probability that the card was drawn from Bag A? [3 points]
(1) $\frac { 1 } { 5 }$
(2) $\frac { 2 } { 9 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 2 } { 7 }$
(5) $\frac { 1 } { 3 }$

For two events $A, B$, $$\mathrm{P}(A \cap B) = \frac{1}{8}, \quad \mathrm{P}\left(B^{C} \mid A\right) = 2\mathrm{P}(B \mid A)$$ what is the value of $\mathrm{P}(A)$? (Here, $B^{C}$ is the complement of $B$.) [3 points]
(1) $\frac{5}{12}$
(2) $\frac{3}{8}$
(3) $\frac{1}{3}$
(4) $\frac{7}{24}$
(5) $\frac{1}{4}$

At a certain school, $60 \%$ of all students commute by bus, and the remaining $40 \%$ walk to school. Of the students who commute by bus, $\frac { 1 } { 20 }$ were late, and of the students who walk, $\frac { 1 } { 15 }$ were late. When one student is randomly selected from all students at this school and is found to be late, what is the probability that this student commuted by bus? [3 points]
(1) $\frac { 3 } { 7 }$
(2) $\frac { 9 } { 20 }$
(3) $\frac { 9 } { 19 }$
(4) $\frac { 1 } { 2 }$
(5) $\frac { 9 } { 17 }$

Among 50 members of a marathon club who participated in a certain marathon, the number of members who completed the marathon and the number who withdrew are as follows. (Unit: persons)

Category	Male	Female
Completed	27	9
Withdrew	8	6

When one member is randomly selected from the participants and is found to be female, the probability that this member completed the marathon is $p$. Find the value of $100 p$. [3 points]

A survey of 320 students at a school regarding membership in the mathematics club found that 60\% of male students and 50\% of female students joined the mathematics club. Let $p _ { 1 }$ be the probability that a randomly selected student from those who joined the mathematics club is male, and let $p _ { 2 }$ be the probability that a randomly selected student from those who joined the mathematics club is female. When $p _ { 1 } = 2 p _ { 2 }$, what is the number of male students at this school? [4 points]
(1) 170
(2) 180
(3) 190
(4) 200
(5) 210

For two events $A$ and $B$, $$\mathrm { P } ( A ) = \frac { 1 } { 3 } , \quad \mathrm { P } ( A \cap B ) = \frac { 1 } { 8 }$$ what is the value of $\mathrm { P } \left( B ^ { C } \mid A \right)$? (Here, $B ^ { C }$ is the complement of $B$.) [4 points]
(1) $\frac { 11 } { 24 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 13 } { 24 }$
(4) $\frac { 7 } { 12 }$
(5) $\frac { 5 } { 8 }$