If $f ^ { \prime } ( x ) > 0$ for all real numbers $x$ and $\int _ { 4 } ^ { 7 } f ( t ) d t = 0$, which of the following could be a table of values for the function $f$ ?
(A)

$x$	$f ( x )$
4	- 4
5	- 3
7	0

(B)

$x$	$f ( x )$
4	- 4
5	- 2
7	5

(C)

$x$	$f ( x )$
4	- 4
5	6
7	3

(D)

$x$	$f ( x )$
4	0
5	0
7	0

(E)

$x$	$f ( x )$
4	0
5	4
7	6

A laboratory studies the spread of a disease in a population. A healthy individual is an individual who has never been affected by the disease. A sick individual is an individual who has been affected by the disease and is not cured. A recovered individual is an individual who has been affected by the disease and has recovered. Once recovered, an individual is immunized and cannot become sick again.
The first observations show that, from one day to the next:

• $5\%$ of individuals become sick;
• $20\%$ of individuals recover.

For every natural number $n$, we denote by $a_n$ the proportion of healthy individuals $n$ days after the start of the experiment, $b_n$ the proportion of sick individuals $n$ days after the start of the experiment, and $c_n$ that of recovered individuals $n$ days after the start of the experiment. We assume that at the start of the experiment, all individuals are healthy, that is $a_0 = 1$, $b_0 = 0$ and $c_0 = 0$.

1. Calculate $a_1$, $b_1$ and $c_1$.
2. a) What is the proportion of healthy individuals who remain healthy from one day to the next? Deduce $a_{n+1}$ as a function of $a_n$. b) Express $b_{n+1}$ as a function of $a_n$ and $b_n$.

We admit that $c_{n+1} = 0.2b_n + c_n$. For every natural number $n$, we define $U_n = \left(\begin{array}{c} a_n \\ b_n \\ c_n \end{array}\right)$ We define the matrices $A = \left(\begin{array}{ccc} 0.95 & 0 & 0 \\ 0.05 & 0.8 & 0 \\ 0 & 0.2 & 1 \end{array}\right)$ and $D = \left(\begin{array}{ccc} 0.95 & 0 & 0 \\ 0 & 0.8 & 0 \\ 0 & 0 & 1 \end{array}\right)$ We admit that there exists an invertible matrix $P$ such that $D = P^{-1} \times A \times P$ and that, for every natural number $n$ greater than or equal to 1, $A^n = P \times D^n \times P^{-1}$.

1. a) Verify that, for every natural number $n$, $U_{n+1} = A \times U_n$. We admit that, for every natural number $n$, $U_n = A^n \times U_0$. b) Prove by induction that, for every non-zero natural number $n$, $$D^n = \left(\begin{array}{ccc} 0.95^n & 0 & 0 \\ 0 & 0.8^n & 0 \\ 0 & 0 & 1 \end{array}\right)$$

We admit that $A^n = \left(\begin{array}{ccc} 0.95^n & 0 & 0 \\ \frac{1}{3}(0.95^n - 0.8^n) & 0.8^n & 0 \\ \frac{1}{3}(3 - 4 \times 0.95^n + 0.8^n) & 1 - 0.8^n & 1 \end{array}\right)$

1. a) Verify that for every natural number $n$, $b_n = \frac{1}{3}(0.95^n - 0.8^n)$ b) Determine the limit of the sequence $(b_n)$. c) We admit that the proportion of sick individuals increases for several days, then decreases. We wish to determine the epidemic peak, that is, the moment when the proportion of sick individuals is at its maximum. To this end, we use the algorithm given in appendix 2 (to be returned with the answer sheet), in which we compare successive terms of the sequence $(b_n)$. Complete the algorithm so that it displays the rank of the day when the epidemic peak is reached and complete the table provided in appendix 2. Conclude.

A chain of stores wishes to build customer loyalty by offering gift vouchers to its privileged customers. Each of them receives a gift voucher of green or red colour on which an amount is written. Gift vouchers are distributed so as to have, in each store, one quarter of red vouchers and three quarters of green vouchers. Green gift vouchers take the value of 30 euros with a probability equal to 0.067 or values between 0 and 15 euros with unspecified probabilities here. Similarly, red gift vouchers take the values 30 or 100 euros with probabilities respectively equal to 0.015 and 0.010 or values between 10 and 20 euros with unspecified probabilities here.
Calculate the probability of having a gift voucher with a value greater than or equal to 30 euros knowing that it is red.

A chain of stores wishes to build customer loyalty by offering gift vouchers to its privileged customers. Each of them receives a gift voucher of green or red colour on which an amount is written. Gift vouchers are distributed so as to have, in each store, one quarter of red vouchers and three quarters of green vouchers. Green gift vouchers take the value of 30 euros with a probability equal to 0.067 or values between 0 and 15 euros with unspecified probabilities here. Similarly, red gift vouchers take the values 30 or 100 euros with probabilities respectively equal to 0.015 and 0.010 or values between 10 and 20 euros with unspecified probabilities here.
Show that an approximate value to $10 ^ { - 3 }$ near of the probability of having a gift voucher with a value greater than or equal to 30 euros is equal to 0.057. For the following question, this value is used.

A chain of stores wishes to build customer loyalty by offering gift vouchers to its privileged customers. Each of them receives a gift voucher of green or red colour on which an amount is written. Gift vouchers are distributed so as to have, in each store, one quarter of red vouchers and three quarters of green vouchers. Green gift vouchers take the value of 30 euros with a probability equal to 0.067 or values between 0 and 15 euros with unspecified probabilities here. Similarly, red gift vouchers take the values 30 or 100 euros with probabilities respectively equal to 0.015 and 0.010 or values between 10 and 20 euros with unspecified probabilities here.
In one of the stores of this chain, out of 200 privileged customers, 6 received a gift voucher with a value greater than or equal to $30 €$.
The manager of the store in question believes that this number is insufficient and doubts the random distribution of gift vouchers in the different stores of the chain. Are his doubts justified?

Exercise 3 (Candidates who have followed the specialization course)

5 POINTS
We have a fair die with 6 faces numbered 1 to 6 and 3 coins A, B and C each having one heads side and one tails side.
A game consists of rolling the die one or more times. After each die roll, if we get 1 or 2, then we flip coin A, if we get 3 or 4, then we flip coin B and if we get 5 or 6, then we flip coin C. At the beginning of the game, all 3 coins are on the tails side.

1. In the algorithm below, 0 codes the tails side and 1 codes the heads side. If $a$ codes one side of coin A, then $1 - a$ codes the other side of coin A.

\begin{verbatim} Variables: a, b, c, d, s are natural integers i, n are integers greater than or equal to 1 Initialization: a takes the value 0 b takes the value 0 c takes the value 0 Input n Processing: For i going from 1 to n do d takes the value of a random integer between 1 and 6 If d <= 2 then a takes the value 1 - a else If d <= 4 then b takes the value 1 - b else c takes the value 1 - c EndIf EndIf s takes the value a + b + c EndFor Output: Display s \end{verbatim}
a. We execute this algorithm by inputting $n = 3$ and assuming that the random values generated successively for $d$ are $1; 4$ and 2. Copy and complete the table given below containing the state of the variables during the execution of the algorithm:

variables	$i$	$d$	$a$	$b$	$c$	$s$
initialization
$1^{\text{st}}$ loop iteration
$2^{\mathrm{nd}}$ loop iteration
$3^{\mathrm{rd}}$ loop iteration

b. Does this algorithm allow us to know whether, after an execution of $n$ rolls, all three coins are on the heads side?
2. For every natural integer $n$, we denote:

• $X_{n}$ the event: ``After $n$ die rolls, all three coins are on the tails side''
• $Y_{n}$ the event: ``After $n$ die rolls, exactly one coin is on the heads side and the others are on the tails side''
• $Z_{n}$ the event: ``After $n$ die rolls, exactly two coins are on the heads side and the other is on the tails side''
• $T_{n}$ the event: ``After $n$ die rolls, all three coins are on the heads side''.

Moreover, we denote $x_{n} = p(X_{n}); y_{n} = p(Y_{n}); z_{n} = p(Z_{n})$ and $t_{n} = p(T_{n})$ the respective probabilities of events $X_{n}, Y_{n}, Z_{n}$ and $T_{n}$.

Exercise 4 — Candidates who have followed the speciality course
We have two urns $U$ and $V$ each containing two balls. Initially, urn $U$ contains two white balls and urn $V$ contains two black balls. We perform successive draws from these urns as follows: each draw consists of taking at random, simultaneously, one ball from each urn and putting it in the other urn. For any non-zero natural number $n$, we denote by $X_n$ the random variable equal to the number of white balls in urn $U$ after $n$ draws.

A rabbit moves in a burrow composed of three galleries, denoted $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$, in each of which it is confronted with a particular stimulus. Each time it is subjected to a stimulus, the rabbit either stays in the gallery where it is or changes gallery. This constitutes a step.
Let $n$ be a natural number. We denote by $a_n$ the probability of the event: ``the rabbit is in gallery A at step $n$''. We denote by $b_n$ the probability of the event: ``the rabbit is in gallery B at step $n$''. We denote by $c_n$ the probability of the event: ``the rabbit is in gallery C at step $n$''.
At step $n = 0$, the rabbit is in gallery A.
A previous study of the rabbit's reactions to different stimuli allows us to model its movements by the following system: $$\left\{ \begin{aligned} a_{n+1} &= \frac{1}{3}a_n + \frac{1}{4}b_n \\ b_{n+1} &= \frac{2}{3}a_n + \frac{1}{2}b_n + \frac{2}{3}c_n \\ c_{n+1} &= \frac{1}{4}b_n + \frac{1}{3}c_n \end{aligned} \right.$$
The objective of this exercise is to estimate in which gallery the rabbit has the greatest probability of being found in the long term.

Part A

Using a spreadsheet, we obtain the following table of values:

	A	B	C	D
1	$n$	$a_n$	$b_n$	$c_n$
2	0	1	0	0
3	1	0,333	0,667	0
4	2	0,278	0,556	0,167
5	3	0,231	0,574	0,194
6	4	0,221	0,571	0,208

It is admitted that $P$ is invertible and that $$P^{-1} = \frac{1}{121}\begin{pmatrix} 120 & 1 \\ -1 & 1 \end{pmatrix}$$
Determine the matrix $D$ defined by $D = P^{-1}AP$.
Prove that, for every natural number $n$, $A^n = PD^nP^{-1}$.
It is admitted henceforth that, for every natural number $n$, $$A^n = \frac{1}{121}\begin{pmatrix} 120 + 0{,}395^n & 1 - 0{,}395^n \\ 120(1 - 0{,}395^n) & 1 + 120 \times 0{,}395^n \end{pmatrix}$$
Deduce an expression for $a_n$ as a function of $n$.
Determine the limit of the sequence $(a_n)$. Conclude.

Part B - Study of a second medium

In this part, we consider a second medium (medium 2), in which we do not know the probability that an atom passes from the excited state to the stable state. We denote by $a$ this probability assumed to be constant. We know, on the other hand, that at each nanosecond, the probability that an atom passes from the stable state to the excited state is 0.01.

1. Give, as a function of $a$, the transition matrix $M$ in medium 2.
2. After a very long time, in medium 2, the proportion of excited atoms stabilizes around $2\%$. It is admitted that there exists a unique vector $X$, called the stationary state, such that $XM = X$, and that $X = (0.98 \quad 0.02)$. Determine the value of $a$.

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

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

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

Exercise 4 — 7 points Theme: Probability An urn contains white and black tokens all indistinguishable to the touch.
A game consists of drawing at random successively and with replacement two tokens from this urn. The following game rule is established:

• a player loses 9 euros if the two tokens drawn are white;
• a player loses 1 euro if the two tokens drawn are black;
• a player wins 5 euros if the two tokens drawn are of different colors.

1. We consider that the urn contains 2 black tokens and 3 white tokens.
   1. [a.] Model the situation using a probability tree.
   2. [b.] Calculate the probability of losing $9\,\text{\euro}$ in one game.
   
   
2. We now consider that the urn contains 3 white tokens and at least two black tokens but we do not know the exact number of black tokens. We will call $N$ the number of black tokens.
   1. [a.] Let $X$ be the random variable giving the gain of the game for one game. Determine the probability distribution of this random variable.
   2. [b.] Solve the inequality for real $x$: $$-x^2 + 30x - 81 > 0$$
   3. [c.] Using the result of the previous question, determine the number of black tokens the urn must contain so that this game is favorable to the player.
   4. [d.] How many black tokens should the player request in order to obtain a maximum average gain?
   
   
3. We observe 10 players who try their luck by playing one game of this game, independently of each other. We assume that 7 black tokens have been placed in the urn (with 3 white tokens). What is the probability of having at least 1 player winning 5 euros?

During a training session, a volleyball player practises serving. The probability that he succeeds on the first serve is equal to 0.85.
We further assume that the following two conditions are satisfied:

• if the player succeeds on a serve, then the probability that he succeeds on the next one is equal to 0.6;
• if the player fails a serve, then the probability that he fails the next one is equal to 0.6.

For any non-zero natural number $n$, we denote by $R _ { n }$ the event ``the player succeeds on the $n$-th serve'' and $\overline { R _ { n } }$ the complementary event.
Part A We are interested in the first two serves of the training session.

1. Represent the situation with a probability tree.
2. Prove that the probability of event $R _ { 2 }$ is equal to 0.57.
3. Given that the player succeeded on the second serve, calculate the probability that he failed the first one.
4. Let $Z$ be the random variable equal to the number of successful serves during the first two serves. a. Determine the probability distribution of $Z$ (you may use the probability tree from question 1). b. Calculate the mathematical expectation $\mathrm { E } ( Z )$ of the random variable $Z$.

Interpret this result in the context of the exercise.
Part B We now consider the general case. For any non-zero natural number $n$, we denote by $x _ { n }$ the probability of event $R _ { n }$.

1. a. Give the conditional probabilities $P _ { R _ { n } } \left( R _ { n + 1 } \right)$ and $P _ { \overline { R _ { n } } } \left( \overline { R _ { n + 1 } } \right)$. b. Show that, for any non-zero natural number $n$, we have: $x _ { n + 1 } = 0.2 x _ { n } + 0.4$.
2. Let the sequence $(u _ { n })$ be defined for any non-zero natural number $n$ by: $u _ { n } = x _ { n } - 0.5$. a. Show that the sequence $(u _ { n })$ is a geometric sequence. b. Determine the expression of $x _ { n }$ as a function of $n$. Deduce the limit of the sequence $\left( x _ { n } \right)$. c. Interpret this limit in the context of the exercise.

6. During the national blood donation week, a blood collection is organized in $N$ French cities chosen at random 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 $M _ { N } = \frac { X _ { 1 } + X _ { 2 } + \cdots + X _ { N } } { N }$. 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 Bienaymé-Chebyshev's inequality allows us to assert that:
$$P \left( 7 < M _ { N } < 7,28 \right) \geq 0,95 .$$

Exercise 2 (6 points)

We consider a function $f$ defined on the interval $] 0 ; + \infty [$. We admit that it is twice differentiable on the interval $] 0 ; + \infty \left[ \right.$. We denote by $f ^ { \prime }$ its derivative function and $f ^ { \prime \prime }$ its second derivative function.
In an orthogonal coordinate system, we have drawn below:

• the representative curve of $f$, denoted $C _ { f }$, on the interval $\left. ] 0 ; 3 \right]$;
• the line $T _ { A }$, tangent to $C _ { f }$ at point $A ( 1 ; 2 )$;
• the line $T _ { B }$, tangent to $C _ { f }$ at point $B ( \mathrm { e } ; \mathrm { e } )$.

We further specify that the tangent $T _ { A }$ passes through point $C ( 3 ; 0 )$. [Figure]

Part A: Graphical readings

We will answer the following questions by justifying them using the graph.

1. Determine the derivative number $f ^ { \prime } ( 1 )$.
2. How many solutions does the equation $f ^ { \prime } ( x ) = 0$ have in the interval $] 0$; 3 ] ?
3. What is the sign of $f ^ { \prime \prime } ( 0,2 )$ ?

Part B: Study of the function $\boldsymbol { f }$

We admit in this part that the function $f$ is defined on the interval $] 0 ; + \infty [$ by:
$$f ( x ) = x \left( 2 ( \ln x ) ^ { 2 } - 3 \ln x + 2 \right)$$
where ln denotes the natural logarithm function.

1. Solve in $\mathbb { R }$ the equation $2 X ^ { 2 } - 3 X + 2 = 0$.

Deduce that $C _ { f }$ does not intersect the x-axis.

The following is a probability distribution table of the random variable $X$.

$X$	$k$	$2 k$	$4 k$	Total
$\mathrm { P } ( X = x )$	$\frac { 4 } { 7 }$	$a$	$b$	1

$\frac { 4 } { 7 } , a , b$ form a geometric sequence in this order, and the mean of $X$ is 24. Find the value of $k$. [3 points]

(Probability and Statistics) A coin is tossed three times, and based on the results, a score is obtained as a random variable $X$ according to the following rules. (가) If the same face does not appear consecutively, the score is 0 points. (나) If the same face appears consecutively exactly twice, the score is 1 point. (다) If the same face appears consecutively three times, the score is 3 points.
What is the variance $\mathrm{V}(X)$ of the random variable $X$? [3 points]
(1) $\frac{9}{8}$
(2) $\frac{19}{16}$
(3) $\frac{5}{4}$
(4) $\frac{21}{16}$
(5) $\frac{11}{8}$

[Probability and Statistics] A certain math class has 10 groups, each consisting of 3 male students and 2 female students. When 2 people are randomly selected from each group, let $X$ be the random variable representing the number of groups in which only male students are selected. What is the expected value $\mathrm { E } ( X )$ of $X$? (Note: No student belongs to more than one group.) [3 points]
(1) 6
(2) 5
(3) 4
(4) 3
(5) 2

The probability distribution table of the random variable $X$ is as follows.

$X$	- 1	0	1	2	Total
$\mathrm { P } ( X = x )$	$\frac { 3 - a } { 8 }$	$\frac { 1 } { 8 }$	$\frac { 3 + a } { 8 }$	$\frac { 1 } { 8 }$	1

When $\mathrm { P } ( 0 \leqq X \leqq 2 ) = \frac { 7 } { 8 }$, what is the value of the expected value $\mathrm { E } ( X )$ of the random variable $X$? [3 points]
(1) $\frac { 1 } { 4 }$
(2) $\frac { 3 } { 8 }$
(3) $\frac { 1 } { 2 }$
(4) $\frac { 5 } { 8 }$
(5) $\frac { 3 } { 4 }$

The probability mass function of a discrete random variable $X$ is $$\mathrm { P } ( X = x ) = \frac { a x + 2 } { 10 } ( x = - 1,0,1,2 )$$ What is the value of the variance $\mathrm { V } ( 3 X + 2 )$ of the random variable $3 X + 2$? (where $a$ is a constant.) [3 points]
(1) 9
(2) 18
(3) 27
(4) 36
(5) 45

The probability distribution of a random variable $X$ is shown in the table below.

$X$	0	1	2	Total
$\mathrm { P } ( X = x )$	$\frac { 1 } { 4 }$	$a$	$2a$	1

What is the value of $\mathrm { E } ( 4X + 10 )$? [3 points]
(1) 11
(2) 12
(3) 13
(4) 14
(5) 15

There is a bag containing 4 white balls and 3 black balls.
Two balls are drawn simultaneously from the bag. If the two balls are of different colors, one coin is flipped 3 times. If the two balls are of the same color, one coin is flipped 2 times. What is the probability that the coin shows heads exactly 2 times in this trial? [3 points]
(1) $\frac { 9 } { 28 }$
(2) $\frac { 19 } { 56 }$
(3) $\frac { 5 } { 14 }$
(4) $\frac { 3 } { 8 }$
(5) $\frac { 11 } { 28 }$

There are 5 drawers, each labeled with a natural number from 1 to 5. Two drawers are randomly assigned to Younghee. Let $X$ be the random variable representing the smaller of the two natural numbers on the assigned drawers. Find the value of $\mathrm { E } ( 10 X )$. [4 points]

A bag contains 1 ball with the number 1, 2 balls with the number 2, and 5 balls with the number 3. One ball is randomly drawn from the bag, the number on the ball is confirmed, and then it is returned. This trial is repeated 2 times. Let $\bar { X }$ be the average of the numbers on the drawn balls. What is the value of $\mathrm { P } ( \bar { X } = 2 )$? [4 points]
(1) $\frac { 5 } { 32 }$
(2) $\frac { 11 } { 64 }$
(3) $\frac { 3 } { 16 }$
(4) $\frac { 13 } { 64 }$
(5) $\frac { 7 } { 32 }$

The probability distribution of a discrete random variable $X$ is shown in the table below.

$X$	- 5	0	5	Total
$\mathrm { P } ( X = x )$	$\frac { 1 } { 5 }$	$\frac { 1 } { 5 }$	$\frac { 3 } { 5 }$	1

Find the value of $\mathrm { E } ( 4 X + 3 )$. [3 points]

A jump is defined as moving from a point $( x , y )$ on the coordinate plane to one of the three points $( x + 1 , y )$, $( x , y + 1 ) , ( x + 1 , y + 1 )$. Let $X$ be the random variable representing the number of jumps when randomly selecting one case from all possible ways to move from point $( 0,0 )$ to point $( 4,3 )$ by repeating jumps. The following is the process of finding the expected value $\mathrm { E } ( X )$ of the random variable $X$. (Here, each case is selected with equal probability.)
Let $N$ be the total number of ways to move from point $( 0,0 )$ to point $( 4,3 )$ by repeating jumps. If the smallest value that the random variable $X$ can take is $k$, then $k =$ (a), and the largest value is $k + 3$.
$$\begin{aligned} & \mathrm { P } ( X = k ) = \frac { 1 } { N } \times \frac { 4 ! } { 3 ! } = \frac { 4 } { N } \\ & \mathrm { P } ( X = k + 1 ) = \frac { 1 } { N } \times \frac { 5 ! } { 2 ! 2 ! } = \frac { 30 } { N } \\ & \mathrm { P } ( X = k + 2 ) = \frac { 1 } { N } \times \text { (b) } \\ & \mathrm { P } ( X = k + 3 ) = \frac { 1 } { N } \times \frac { 7 ! } { 3 ! 4 ! } = \frac { 35 } { N } \end{aligned}$$
and $$\sum _ { i = k } ^ { k + 3 } \mathrm { P } ( X = i ) = 1$$ so $N =$ (c). Therefore, the expected value $\mathrm { E } ( X )$ of the random variable $X$ is as follows. $$\mathrm { E } ( X ) = \sum _ { i = k } ^ { k + 3 } \{ i \times \mathrm { P } ( X = i ) \} = \frac { 257 } { 43 }$$
When the numbers corresponding to (a), (b), (c) are $a , b , c$ respectively, what is the value of $a + b + c$? [4 points]
(1) 190
(2) 193
(3) 196
(4) 199
(5) 202

A jump is defined as moving from a point $( x , y )$ on the coordinate plane to one of the three points $( x + 1 , y )$, $( x , y + 1 )$, or $( x + 1 , y + 1 )$. Let $X$ be the random variable representing the number of jumps that occur when one case is randomly selected from all cases of moving from point $( 0,0 )$ to point $( 4,3 )$ by repeating jumps. The following is the process of finding the mean $\mathrm { E } ( X )$ of the random variable $X$. (Here, each case is selected with equal probability.)
Let $N$ be the total number of cases of moving from point $( 0,0 )$ to point $( 4,3 )$ by repeating jumps. If the smallest value that the random variable $X$ can take is $k$, then $k =$ (가), and the largest value is $k + 3$.
$$\begin{aligned} & \mathrm { P } ( X = k ) = \frac { 1 } { N } \times \frac { 4 ! } { 3 ! } = \frac { 4 } { N } \\ & \mathrm { P } ( X = k + 1 ) = \frac { 1 } { N } \times \frac { 5 ! } { 2 ! 2 ! } = \frac { 30 } { N } \\ & \mathrm { P } ( X = k + 2 ) = \frac { 1 } { N } \times \text { (나) } \\ & \mathrm { P } ( X = k + 3 ) = \frac { 1 } { N } \times \frac { 7 ! } { 3 ! 4 ! } = \frac { 35 } { N } \end{aligned}$$
and $$\sum _ { i = k } ^ { k + 3 } \mathrm { P } ( X = i ) = 1$$ so $N =$ (다). Therefore, the mean $\mathrm { E } ( X )$ of the random variable $X$ is as follows. $$\mathrm { E } ( X ) = \sum _ { i = k } ^ { k + 3 } \{ i \times \mathrm { P } ( X = i ) \} = \frac { 257 } { 43 }$$
When the numbers that fit (가), (나), and (다) are $a$, $b$, and $c$, respectively, what is the value of $a + b + c$? [4 points]
(1) 190
(2) 193
(3) 196
(4) 199
(5) 202

The probability distribution of the random variable $X$ is shown in the table below.

$X$	0.121	0.221	0.321	Total
$\mathrm { P } ( X = x )$	$a$	$b$	$\frac { 2 } { 3 }$	1

The following is the process of finding $\mathrm { V } ( X )$ when $\mathrm { E } ( X ) = 0.271$. Let $Y = 10 X - 2.21$. The probability distribution of the random variable $Y$ is shown in the table below.

$Y$	$-1$	0	1	Total
$\mathrm { P } ( Y = y )$	$a$	$b$	$\frac { 2 } { 3 }$	1

Since $\mathrm { E } ( Y ) = 10 \mathrm { E } ( X ) - 2.21 = 0.5$, $a =$ (가), $b =$ (나) and $\mathrm { V } ( Y ) = \frac { 7 } { 12 }$. On the other hand, since $Y = 10 X - 2.21$, we have $\mathrm { V } ( Y ) =$ (다) $\times \mathrm { V } ( X )$. Therefore, $\mathrm { V } ( X ) = \frac { 1 } { \text{(다)} } \times \frac { 7 } { 12 }$. When the values in (가), (나), and (다) are $p$, $q$, and $r$ respectively, find the value of $pqr$. (Here, $a$ and $b$ are constants.) [4 points]
(1) $\frac { 13 } { 9 }$
(2) $\frac { 16 } { 9 }$
(3) $\frac { 19 } { 9 }$
(4) $\frac { 22 } { 9 }$
(5) $\frac { 25 } { 9 }$