A factory produces products that are sold with 50 items per box. The number of defective items in a box follows a binomial distribution with mean $m$ and variance $\frac { 48 } { 25 }$. Before selling a box, all 50 products are inspected to find defective items, which costs a total of 60,000 won. If a box is sold without inspection, an after-sales service cost of $a$ won is required for each defective item.
When the expected value of the cost of inspecting all products in a box equals the expected cost of after-sales service, find the value of $\frac { a } { 1000 }$. (Given that $a$ is a constant and $m$ is a natural number not exceeding 5.) [4 points]

18. (This question is worth 12 points)
A shopping mall is holding a promotional lottery. After purchasing goods of a certain amount, customers can participate in a lottery. Each lottery involves randomly drawing one ball from box A (containing 4 red balls and 6 white balls) and one ball from box B (containing 5 red balls and 5 white balls). If both balls drawn are red, the customer wins the first prize; if exactly one ball is red, the customer wins the second prize; if neither ball is red, the customer wins no prize.
(1) Find the probability that a customer wins a prize in one lottery;
(2) If a customer has 3 lottery chances, let X denote the number of times the customer wins the first prize in the 3 lotteries. Find the probability distribution and mathematical expectation of X.

13. The defect rate of a batch of products is 0.02. Drawing one item at a time with replacement from this batch, 100 times total, let $X$ denote the number of defective items drawn. Then $D X = $ ______

A shooting competition is divided into two stages. Each participating team consists of two members. The specific rules are as follows: In the first stage, one team member shoots 3 times. If all 3 shots miss, the team is eliminated with a score of 0. If at least one shot is made, the team advances to the second stage, where the other team member shoots 3 times, earning 5 points for each made shot and 0 points for each missed shot. The team's final score is the total points from the second stage. A participating team consists of members A and B. Let the probability that A makes each shot be $p$, and the probability that B makes each shot be $q$. Each shot is independent.
(1) If $p = 0.4$ and $q = 0.5$, with A participating in the first stage, find the probability that the team's score is at least 5 points.
(2) Assume $0 < p < q$.
(i) To maximize the probability that the team's score is 15 points, who should participate in the first stage?
(ii) To maximize the expected value of the team's score, who should participate in the first stage?

Let $x = (x_0, \ldots, x_{n-1}) \in \{0,1\}^n$ and consider a noisy observation $$O(x) = \left(O_0(x), O_1(x), \ldots, O_{n-1}(x)\right)$$ of source $x$.
a. If $0 \leq j \leq k \leq n-1$, show that $\mathbb{P}\left(N \geq j \text{ and } I_j = k\right) = p\binom{k}{j}p^j(1-p)^{k-j}$.
b. Show that, for all $0 \leq j \leq n-1$, $$\mathbb{E}\left[O_j(x)\right] = p\sum_{k=j}^{n-1} x_k \binom{k}{j} p^j (1-p)^{k-j}.$$
c. Show that for all $\omega \in \mathbb{C}$, $$\mathbb{E}\left[\sum_{j=0}^{n-1} O_j(x) \omega^j\right] = p \sum_{k=0}^{n-1} x_k (p\omega + 1 - p)^k.$$

A message is transmitted over a channel where each bit is inverted independently with probability $1-p \in ]0,1[$. Blocks of $r$ bits are transmitted. $X$ denotes the number of inversions during the transmission of a block of $r$ bits. We consider $\alpha \in ]0,1[$ and the condition $$\mathbb{P}(X \geqslant 2) \leqslant 1 - \alpha. \tag{II.2}$$ We set $a = \frac{p \ln(p)}{p-1}$ and $x = r\ln(p) - a$. Prove that condition (II.2) is equivalent to the condition $$x \mathrm{e}^{x} \leqslant -\alpha a \mathrm{e}^{-a}.$$

A message is transmitted over a channel where each bit is inverted independently with probability $1-p \in ]0,1[$. Blocks of $r$ bits are transmitted. $X$ denotes the number of inversions during the transmission of a block of $r$ bits. We consider $\alpha \in ]0,1[$ and the condition $$\mathbb{P}(X \geqslant 2) \leqslant 1 - \alpha. \tag{II.2}$$ With $a = \frac{p\ln(p)}{p-1}$ and $x = r\ln(p) - a$, condition (II.2) is equivalent to $xe^x \leqslant -\alpha a e^{-a}$. Let $V$ and $W$ be the Lambert functions defined in Part I. Using one of the functions $V$ and $W$ and Question 10, study the existence of a largest natural integer $r$ satisfying condition (II.2).

What is the distribution of $Y _ { n }$ ? Deduce the expectation and variance of $Y _ { n }$.

Let $n \in \mathbb { N } ^ { * }$ and let $\gamma = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 n } \right)$ be a Dyck path of length $2 n$. For $t \in \mathbb { N }$, express $\mathbb { P } \left( A _ { t , \gamma } \right)$ as a function of $n$ and $p$, where $A _ { t , \gamma } = \bigcap _ { k = 1 } ^ { m } \left( X _ { t + k } = \gamma _ { k } \right)$.

Let $( \Omega , \mathcal { A } , \mathbb { P } )$ be a probability space and $X$ a discrete random variable such that $\mathbb { P } ( X = - 1 ) = 1 / 2$ and $\mathbb { P } ( X = 1 ) = 1 / 2$. Consider a sequence $\left( X _ { i } \right) _ { i \in \mathbb { N } ^ { * } }$ of mutually independent discrete random variables with the same distribution as $X$. Set $S _ { 0 } = 0$ and $S _ { n } = \sum _ { i = 1 } ^ { n } X _ { i }$. Set $Y _ { i } = \frac { X _ { i } + 1 } { 2 }$ and $T _ { n } = \sum _ { i = 1 } ^ { n } Y _ { i }$. For all $n \in \mathbb { N } ^ { * }$ and all $k \in \llbracket 0 , n \rrbracket$, set $x _ { n , k } = - \sqrt { n } + \frac { 2 k } { \sqrt { n } }$. The function $B_n$ is defined as in Q19.
Show that, for all $j \in \llbracket 0 , n \rrbracket$, $$\mathbb { P } \left( \left\{ T _ { n } = j \right\} \right) = \int _ { x _ { n , j } - 1 / \sqrt { n } } ^ { x _ { n , j } + 1 / \sqrt { n } } B _ { n } ( x ) \mathrm { d } x$$

If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is
(1) $22 \left( \frac { 1 } { 3 } \right) ^ { 11 }$
(2) $\frac { 5 } { 19 }$
(3) $55 \left( \frac { 2 } { 3 } \right) ^ { 10 }$
(4) $220 \left( \frac { 1 } { 3 } \right) ^ { 12 }$

A six faced die is biased such that $3 \times P ($ a prime number $) = 6 \times P ($ a composite number $) = 2 \times P ( 1 )$. Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is
(1) $\frac { 3 } { 11 }$
(2) $\frac { 5 } { 11 }$
(3) $\frac { 7 } { 11 }$
(4) $\frac { 8 } { 11 }$

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 certain online game offers a ``ten-draw'' card-pulling mechanism. ``Ten-draw'' means the system automatically performs ten card-pulling actions. If each ``ten-draw'' requires 1500 tokens, the probability of drawing a gold card is 2\% for the first nine draws and 10\% for the tenth draw. A certain student has 23000 tokens and continuously uses ``ten-draw'' until unable to draw anymore. The expected value of the number of gold cards drawn is (13-1)(13-2) cards.

Problem 6
Consider $n$ random variables $X _ { 1 } , X _ { 2 } , \cdots , X _ { n }$ that can take the values 0 and 1. Here, $n$ is an integer greater than or equal to 4. The probability of an event $A$ is denoted by $P ( A )$, and the conditional probability of the event $A$ given an event $B$ is denoted by $P ( A \mid B )$. The intersection between the event $A$ and the event $B$ is denoted by $A \wedge B$. Answer the following questions.
I. Let us assume that the $X _ { 1 } , X _ { 2 } , \cdots , X _ { n }$ are independent. In addition, assume that each $X _ { k } \quad ( k = 1, 2 , \cdots , n )$ takes the value 1 with the probability $p$ and the value 0 with the probability $1 - p$, i.e., $P \left( X _ { k } = 1 \right) = p$ and $P \left( X _ { k } = 0 \right) = 1 - p$.

1. Find the expected value and the variance of the sum of the $X _ { 1 } , X _ { 2 } , \cdots , X _ { n }$.
2. The random variables $X _ { 1 } , X _ { 2 } , \cdots , X _ { n }$ are arranged in the row $X _ { n } \cdots X _ { 2 } X _ { 1 }$. Let $Y$ be the integer value obtained by regarding that row as an $n$-digit binary number. For example, in the case that $n = 4$, $Y = 5$ when the row $X _ { 4 } X _ { 3 } X _ { 2 } X _ { 1 }$ is 0101, and $Y = 13$ when the row $X _ { 4 } X _ { 3 } X _ { 2 } X _ { 1 }$ is 1101. $Y$ is a random variable that takes integer values from 0 to $2 ^ { n } - 1$. Obtain the expected value and variance of $Y$.

II. The values of the random variables $X _ { 1 } , X _ { 2 } , \cdots , X _ { n }$ are obtained sequentially according to the following steps. First, $X _ { 1 }$ takes the value 1 with the probability $p$ and the value 0 with the probability $1 - p$. Then, $X _ { k } \quad ( k = 2, 3 , \cdots , n )$ takes the same value as $X _ { k - 1 }$ with the probability $q$ and the value different from $X _ { k - 1 }$ with the probability $1 - q$, i.e., $P \left( X _ { k } = 1 \mid X _ { k - 1 } = 1 \right) = P \left( X _ { k } = 0 \mid X _ { k - 1 } = 0 \right) = q$ and $P \left( X _ { k } = 1 \mid X _ { k - 1 } = 0 \right) = P \left( X _ { k } = 0 \mid X _ { k - 1 } = 1 \right) = 1 - q$.

1. Let $P \left( X _ { k } = 1 \right)$ be represented by $r _ { k }$, where $k$ is an integer varying from 1 to $n$. Derive a recurrence equation for $r _ { k }$. Solve this recurrence equation to express $r _ { k }$ with $p , q$, and $k$.
2. Obtain the probability $P \left( X _ { 1 } = 1 \wedge X _ { 2 } = 0 \wedge X _ { 3 } = 1 \wedge X _ { 4 } = 0 \right)$.
3. Obtain the probability $P \left( X _ { 3 } = 1 \mid X _ { 1 } = 0 \wedge X _ { 2 } = 1 \wedge X _ { 4 } = 1 \right)$.