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

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

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

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

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

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

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

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

Part A:

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

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

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

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

PART A

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

PART B

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

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

PART C

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

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

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

Stéphanie is practising shooting. We have the following data:

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

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

Exercise 1 — 6 points

Main topics covered: Probability

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

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

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

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

We interview a skier at random and consider the events:

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

The two parts can be worked on independently

Part A

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

Part B

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

We consider the following events:

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

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

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

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

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

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

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

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

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

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

Part B

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

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 $q_n \geqslant 0.9999$.

The director of a school wishes to conduct a study among students who took the final examination to analyze how they think they performed on this exam. For this study, students are asked at the end of the exam to answer individually the question: ``Do you think you passed the exam?''.
Only the answers ``yes'' or ``no'' are possible, and it is observed that $91.7\%$ of the students surveyed answered ``yes''. Following the publication of exam results, it is discovered that:

• $65\%$ of students who failed answered ``no'';
• $98\%$ of students who passed answered ``yes''.

A student who took the exam is randomly selected. We denote by $R$ the event ``the student passed the exam'' and $Q$ the event ``the student answered ``yes'' to the question''. For any event $A$, we denote by $P(A)$ its probability and $\bar{A}$ its complementary event.
Throughout the exercise, probabilities are, if necessary, rounded to $10^{-3}$ near.

1. Specify the values of the probabilities $P(Q)$ and $P_{\bar{R}}(\bar{Q})$.
2. Let $x$ be the probability that the randomly selected student passed the exam. a. Copy and complete the weighted tree below. b. Show that $x = 0.9$.
3. The student selected answered ``yes'' to the question. What is the probability that he passed the exam?
4. The grade obtained by a randomly selected student is an integer between 0 and 20. It is assumed to be modeled by a random variable $N$ that follows the binomial distribution with parameters $(20; 0.615)$.
   The director wishes to award a prize to students with the best results.
   Starting from which grade should she award prizes so that $65\%$ of students are rewarded?
5. Ten students are randomly selected.
   The random variables $N_1, N_2, \ldots, N_{10}$ model the grade out of 20 obtained on the exam by each of them. It is admitted that these variables are independent and follow the same binomial distribution with parameters $(20; 0.615)$. Let $S$ be the variable defined by $S = N_1 + N_2 + \cdots + N_{10}$. Calculate the expectation $E(S)$ and the variance $V(S)$ of the random variable $S$.
6. Consider the random variable $M = \frac{S}{10}$. a. What does this random variable $M$ model in the context of the exercise? b. Justify that $E(M) = 12.3$ and $V(M) = 0.47355$. c. Using the Bienaymé-Chebyshev inequality, justify the statement below. ``The probability that the average grade of ten randomly selected students is strictly between 10.3 and 14.3 is at least $80\%$''.

Exercise 1 — Part A
The centre offers people coming for a weekend an introductory roller skating formula consisting of two training sessions. We randomly choose a person among those who have subscribed to this formula. We denote by $A$ and $B$ the following events:

• A: ``The person falls during the first session'';
• B: ``The person falls during the second session''.

For any event $E$, we denote $P(E)$ its probability and $\bar{E}$ its complementary event. Observations allow us to assume that $P(A) = 0{,}6$. Furthermore, we observe that:

• If the person falls during the first session, the probability that they fall during the second is 0.3;
• If the person does not fall during the first session, the probability that they fall during the second is 0.4.

1. Represent the situation with a probability tree.
2. Calculate the probability $P(\bar{A} \cap \bar{B})$ and interpret the result.
3. Show that $P(B) = 0{,}34$.
4. The person does not fall during the second training session. Calculate the probability that they did not fall during the first session.
5. We call $X$ the random variable which, for each sample of 100 people who have subscribed to the formula, associates the number of them who did not fall during either the first or the second session. We assimilate the choice of a sample of 100 people to a draw with replacement. We admit that the probability that a person does not fall during either the first or the second session is 0.24.
   1. [a.] Show that the random variable $X$ follows a binomial distribution whose parameters you will specify.
   2. [b.] What is the probability of having, in a sample of 100 people who have subscribed to the formula, at least 20 people who do not fall during either the first or the second session?
   3. [c.] Calculate the expectation $E(X)$ and interpret the result in the context of the exercise.

A coin is tossed 7 times. What is the probability of satisfying the following conditions? [4 points] (가) Heads appears at least 3 times. (나) There is a case where heads appears consecutively.
(1) $\frac { 11 } { 16 }$
(2) $\frac { 23 } { 32 }$
(3) $\frac { 3 } { 4 }$
(4) $\frac { 25 } { 32 }$
(5) $\frac { 13 } { 16 }$

During the COVID-19 pandemic prevention and control period, a supermarket opened online sales services and can complete 1200 orders per day. Due to a sharp increase in order volume, orders have accumulated. To solve this problem, many volunteers eagerly signed up to help with order fulfillment. It is known that the supermarket had 500 accumulated unfulfilled orders on a certain day, and the probability that the next day's new orders exceed 1600 is 0.05. Each volunteer can complete 50 orders per day. To ensure that the probability of completing accumulated orders and current day orders within two days is at least 0.95, the minimum number of volunteers needed is
A． 10 people
B． 18 people
C． 24 people
D． 32 people

Two people, A and B, practice table tennis. The winner of each ball scores 1 point, the loser scores 0 points. Let the probability that A wins each ball be $p$ $\left(\frac{1}{2} < p < 1\right)$, the probability that B wins be $q$, with $p + q = 1$. The outcome of each ball is independent. For a positive integer $k \geq 2$, let $p_k$ denote the probability that after $k$ balls, A has scored at least 2 more points than B, and let $q_k$ denote the probability that after $k$ balls, B has scored at least 2 more points than A.
(1) Find $p_3, p_4$ (expressed in terms of $p$).
(2) If $\frac{p_4 - p_3}{q_4 - q_3} = 4$, find $p$.
(3) Prove: For any positive integer $m$, $p_{2m+1} - q_{2m+1} < p_{2m} - q_{2m} < p_{2m+2} - q_{2m+2}$.

Let $C _ { 1 }$ and $C _ { 2 }$ be two biased coins such that the probabilities of getting head in a single toss are $\frac { 2 } { 3 }$ and $\frac { 1 } { 3 }$, respectively. Suppose $\alpha$ is the number of heads that appear when $C _ { 1 }$ is tossed twice, independently, and suppose $\beta$ is the number of heads that appear when $C _ { 2 }$ is tossed twice, independently. Then the probability that the roots of the quadratic polynomial $x ^ { 2 } - \alpha x + \beta$ are real and equal, is
(A) $\frac { 40 } { 81 }$
(B) $\frac { 20 } { 81 }$
(C) $\frac { 1 } { 2 }$
(D) $\frac { 1 } { 4 }$

For a game, each of two people, A and B , has a bag containing three cards on which the numbers 1, 2, and 3 are written, each number on a different card. In the game, A and B each take out one card from their own bag and compare the numbers. If the numbers are the same, the game is a draw. If the numbers are different, the person with the greater number wins.
(1) For a single game the probability of a draw is $\frac { \mathbf { L } } { \mathbf { M } }$.
(2) If this game is successively played four times, replacing the cards after each game, let us find the probabilities for the following.
(i) The probability that A wins three times or more is $\frac { \mathbf { N } } { \mathbf{O} }$.
(ii) The probability that A wins once and loses once and two games are draws is $\frac { \mathrm { P } } { \mathrm { QR } }$.
(iii) The probability that the number of times that A wins and the number of times that B wins are the same is $\frac { \mathbf { S T } } { \mathbf { U V } }$. Hence, the probability that the number of times that A wins is greater than the number of times that B wins is $\frac { \mathbf { W } \mathbf { X } } { \mathbf{UV} }$.

A biased coin has probability $\frac { 1 } { 3 }$ of showing heads and probability $\frac { 2 } { 3 }$ of showing tails. On a coordinate plane, a game piece moves to the next position based on the result of flipping this coin, according to the following rules: (I) If heads appears, the piece moves from its current position in the direction and distance of vector $( - 1,2 )$ to the next position; (II) If tails appears, the piece moves from its current position in the direction and distance of vector $( 1,0 )$ to the next position. For example: If the game piece is currently at coordinates $( 2,4 )$ and tails appears, the piece moves to coordinates $( 3,4 )$. Suppose the game piece starts at the origin $( 0,0 )$ and, according to the above rules, flips the coin 6 times consecutively, with each flip being independent. After 6 moves, the game piece is most likely to stop at coordinates (12), (13)).

It is desired to place 4 identical chess rooks on a $5 \times 5$ chessboard. Since rooks can capture pieces in the same row or column, the placement rule is that at most one rook can be placed in each row and each column. Given that rooks are not placed in the first, third, and fifth squares of the first row (as shown by the crossed squares in the diagram), how many ways are there to place the rooks?
(1) 216
(2) 240
(3) 288
(4) 312
(5) 360

A store launches a lottery activity offering four different styles of fruit figurines as prizes. Each lottery draw yields 1 figurine, and each style has an equal probability of being drawn. A person decides to draw 4 times. What is the probability that he draws exactly 3 different styles of figurines?
(1) $\frac { 5 } { 16 }$
(2) $\frac { 3 } { 8 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 9 } { 16 }$
(5) $\frac { 5 } { 8 }$