Three defective oranges are accidently mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denote the number of defective oranges, then the variance of $x$ is
(1) $28/75$
(2) $18/25$
(3) $26/75$
(4) $14/25$

A coin is tossed three times. Let $X$ denote the number of times a tail follows a head. If $\mu$ and $\sigma ^ { 2 }$ denote the mean and variance of $X$, then the value of $64 \left( \mu + \sigma ^ { 2 } \right)$ is:
(1) 51
(2) 64
(3) 32
(4) 48

There are two boxes, A and B.
In box A, there are three cards on which the number 0 is written, two cards on which the number 2 is written, and one card on which the number 3 is written.
In box B, there are two cards on which the number 1 is written, and three cards on which the number 2 is written.
Take two cards together from box A and one card from box B. Denote the product of the numbers on the three cards by $X$.
The total number of values which $X$ can take is A. The maximum value which $X$ can take is $\mathbf{BC}$. The minimum value which $X$ can take is D.
The probability that $X = \mathrm{BC}$ is $\frac{\mathbf{E}}{\mathrm{F}}$, and the probability that $X = \square\mathrm{D}$ is $\frac{\mathbf{H}}{\mathbf{I}}$.

In a bag there are a total of nine balls: one white, three red and five black. The white ball is worth five points, a red ball is worth three points, and a black ball is worth one point. Two balls are taken from the bag together, and the trial is scored by the sum of the values of the two balls.
(1) The highest possible score is $\mathbf{A}$, and the probability that it happens is $\dfrac{\mathbf{B}}{\mathbf{CD}}$.
(2) The probability that the score is 6 is $\dfrac{\mathbf{E}}{\mathbf{F}}$.
(3) The expected value of the score is $\dfrac{\mathbf{GH}}{\mathbf{I}}$.

Let us throw one dice three times, and let the number that comes up on the first throw be $a$, on the second throw be $b$, and on the third throw be $c$. Using these $a , b$ and $c$, we consider the quadratic function $f ( x ) = a x ^ { 2 } + b x + c$.
(1) The probability that $b = 4$ and that the quadratic equation $f ( x ) = 0$ has two different real solutions is $\frac { \mathbf { N } } { \mathbf { O P Q } }$.
(2) Let us find the probability that $f ( 10 ) > 453$.
The number of the cases of $( a , b , c )$ such that $f ( 10 ) > 453$ is as follows: when $a = 4$ and $b = 5$, it is $\mathbf { R }$; when $a = 4$ and $b = 6$, it is $\mathbf{S}$; when $a = 5$, it is $\mathbf { T U }$; when $a = 6$, it is $\mathbf{VW}$. Hence, the probability that $f ( 10 ) > 453$ is $\frac { \mathbf { X } } { \mathbf { Y } }$.

Let us throw one dice three times, and let the number that comes up on the first throw be $a$, on the second throw be $b$, and on the third throw be $c$. Using these $a , b$ and $c$, we consider the quadratic function $f ( x ) = a x ^ { 2 } + b x + c$.
(1) The probability that $b = 4$ and that the quadratic equation $f ( x ) = 0$ has two different real solutions is $\frac { \mathbf { N } } { \mathbf { O } \mathbf { P Q } }$.
(2) Let us find the probability that $f ( 10 ) > 453$.
The number of the cases of $( a , b , c )$ such that $f ( 10 ) > 453$ is as follows: when $a = 4$ and $b = 5$, it is $\mathbf { R }$; when $a = 4$ and $b = 6$, it is $\mathbf{S}$; when $a = 5$, it is $\mathbf{TU}$; when $a = 6$, it is $\mathbf{VW}$. Hence, the probability that $f ( 10 ) > 453$ is $\frac { \mathbf { X } } { \mathbf { Y } }$.

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 $\frac { \mathbf { W } \mathbf { X } } { \mathbf{UV} }$.

If the possible values of random variable $X$ are $1 , 2 , 3 , 4$ , and the probability $P ( X = k )$ is proportional to $\frac { 1 } { k }$ , then the probability $P ( X = 3 )$ is $\frac { \text{(9)(11)} } { \text{(10)(11)} }$ . (Reduce to lowest terms)

There are two boxes $A$ and $B$. Box $A$ contains 6 white balls and 4 red balls, and box $B$ contains 8 white balls and 2 blue balls. There are three lottery methods (each ball in each box has equal probability of being drawn): (I) First draw one ball from box $A$; if a red ball is drawn, stop; if a white ball is drawn, then draw one ball from box $B$; (II) First draw one ball from box $B$; if a blue ball is drawn, stop; if a white ball is drawn, then draw one ball from box $A$; (III) Simultaneously draw one ball from each of boxes $A$ and $B$. The prize rules are: Among red and blue balls, if only a red ball is drawn, win 50 yuan; if only a blue ball is drawn, win 100 yuan; if both colors are drawn, still win only 100 yuan; if neither color is drawn, win nothing. Let $E_{1}$, $E_{2}$, $E_{3}$ denote the expected values of winnings for methods (I), (II), (III) respectively. Select the correct option.
(1) $E_{1} > E_{2} > E_{3}$
(2) $E_{1} = E_{2} > E_{3}$
(3) $E_{2} = E_{3} > E_{1}$
(4) $E_{1} = E_{3} > E_{2}$
(5) $E_{3} > E_{2} > E_{1}$

There is a game involving moving a game piece on a number line. The way to move the piece is determined by rolling a fair die, with the following rules: (I) When the die shows 1 point, the piece does not move. (II) When the die shows 3 or 5 points, the piece moves left (negative direction) by ``that point number minus 1'' units. (III) When the die shows an even number, the piece moves right (positive direction) by ``half of that point number'' units. On the first die roll, the piece starts at the origin. From the second roll onwards, the piece starts from the position it was in after the previous roll. For example, if two die rolls result in 5 points and 2 points respectively, the piece first moves left 4 units to coordinate $-4$, then moves right 1 unit to coordinate $-3$. Select the correct options.
(1) After rolling the die once, the probability that the piece is at distance 2 from the origin is $\frac { 1 } { 2 }$
(2) After rolling the die once, the expected value of the piece's coordinate is 0
(3) After rolling the die twice, the piece's coordinate could be $-5$
(4) After rolling the die twice, given that the sum of the two rolls is odd, the probability that the piece's coordinate is positive is $\frac { 4 } { 9 }$
(5) After rolling the die three times, the probability that the piece is at the origin is $\left( \frac { 1 } { 6 } \right) ^ { 3 }$

A company holds a year-end lottery. Each person randomly draws two cards from six cards numbered 1 to 6. Assume each card has an equal chance of being drawn, and the rules are as follows: (I) If the sum of the numbers on the two cards is odd, the person wins 100 yuan and the lottery ends; (II) If the sum is even, the two cards are discarded, and two cards are randomly drawn from the remaining four cards. If the sum of their numbers is odd, the person wins 50 yuan; otherwise, there is no prize and the lottery ends. According to the above rules, what is the expected value of the prize money for each person participating in this lottery?
(1) 50
(2) 70
(3) 72
(4) 80
(5) 100

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.

There is a circular clock with numbers $1, 2 , \ldots , 12$ marked in clockwise order (as shown in the figure). Initially, a game piece is placed at the ``12'' o'clock position on this clock. Then, each time a fair coin is tossed, the game piece is moved according to the following rules:

• If heads appears, move the game piece 5 clock positions clockwise from its current position.
• If tails appears, move the game piece 5 clock positions counterclockwise from its current position.

For example: If the coin is tossed three times and all are heads, the game piece moves to the ``5'' o'clock position on the first move, the ``10'' o'clock position on the second move, and the ``3'' o'clock position on the third move.
For any positive integer $n$, let the random variable $X _ { n }$ represent the clock position of the game piece after $n$ moves according to the above rules, $P \left( X _ { n } = k \right)$ represents the probability that $X _ { n } = k$ (where $k = 1, 2 , \ldots , 12$), and let $E \left( X _ { n } \right)$ represent the expected value of $X _ { n }$. Select the correct options.
(1) $E \left( X _ { 1 } \right) = 6$
(2) $P \left( X _ { 2 } = 12 \right) = \frac { 1 } { 4 }$
(3) $P \left( X _ { 8 } = 5 \right) \geq \frac { 1 } { 2 ^ { 8 } }$
(4) $P \left( X _ { 8 } = 4 \right) = P \left( X _ { 8 } = 8 \right)$
(5) $E \left( X _ { 8 } \right) \leq 7$

A lottery game has a single-play winning probability of 0.1, and each play is an independent event. For each positive integer $n$, let $p_{n}$ be the probability of winning at least once in $n$ plays of this game. Select the correct options.
(1) $p_{n+1} > p_{n}$
(2) $p_{3} = 0.3$
(3) $\langle p_{n} \rangle$ is an arithmetic sequence
(4) Playing this game two or more times, the probability of not winning on the first play and winning on the second play equals $p_{2} - p_{1}$
(5) When playing this game $n$ times with $n \geq 2$, the probability of winning at least 2 times equals $2p_{n}$

Xiaoming wrote a program to move a robot on a $2 \times 2$ chessboard, as shown in the figure. Each execution, the program selects one direction from ``up, down, left, right'' with equal probability for each direction, and instructs the robot to move one square in that direction. However, if the selected direction would move the robot off the board, the robot stays in place. Each execution starts from the robot's current position with a newly selected direction by the program. Assume the robot's initial position is $A$. Let $a _ { n } , b _ { n } , c _ { n }$, and $d _ { n }$ be the probabilities that the robot is at positions $A , B , C , D$ respectively after executing the program $n$ times. Select the correct options.
(1) $b _ { 1 } = \frac { 1 } { 4 }$
(2) $b _ { 2 } = \frac { 1 } { 8 }$
(3) $a _ { 2 } + d _ { 2 } = \frac { 3 } { 4 }$

$A$	$B$
$C$	$D$

(4) $b _ { 99 } = c _ { 99 }$
(5) $a _ { 100 } + d _ { 100 } > \frac { 1 } { 2 }$

A shopping mall holds a raffle drawing activity with on-site registration. After registration closes, the host places the same number of raffle balls as the number of registrants, of which 10 balls are marked as lucky prizes: 5 balls for a 5000 yuan gift voucher and 5 balls for an 8000 yuan gift voucher. Each ball has an equal probability of being drawn, and balls are not replaced after drawing. Before the drawing, the organizers announce a winning probability of 0.4\% based on the number of prizes and registrants. After the drawing begins, each person draws a ball in order, and each person has only one chance to draw. If among the first 100 participants, exactly 1 person draws a 5000 yuan voucher and no one draws an 8000 yuan voucher, then the expected value of the prize amount that the 101st person can receive is (15－1)(15－2) yuan.

There is a card-drawing prize activity with the following rules: In an opaque box, there are 2 cards marked with ``1000 yuan'' and 3 cards marked with ``0 yuan''. A participant randomly draws one card from the box. Without knowing the amount marked on the drawn card, the host then places a card marked with ``500 yuan'' into the box. At this point, the participant has two choices: (I) Keep the originally drawn card; the amount marked on that card is the prize won. (II) Discard the originally drawn card without returning it, and randomly draw another card from the box; the amount marked on that card is the prize won.
A participant joins this activity. Assume each card has an equal chance of being drawn. Select the correct options.
(1) If the participant chooses (I), the probability of winning 0 yuan is $\frac { 3 } { 5 }$
(2) If the participant chooses (I), the expected value of the prize is 500 yuan
(3) If the participant chooses (II), the probability of winning 1000 yuan is $\frac { 2 } { 5 }$
(4) If the participant chooses (II), the probability of winning 0 yuan is $\frac { 12 } { 25 }$
(5) If the participant chooses (II), the expected value of the prize is 420 yuan

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
If using Method 1 to participate in the lottery, what is the probability of paying a total of 300 yuan to obtain one action figure? (Single choice question, 2 points)
(1) $\left(\frac{2}{5}\right)^{2}$
(2) $\left(\frac{2}{5}\right)^{3}$
(3) $\left(\frac{3}{5}\right)^{2}$
(4) $\left(\frac{3}{5}\right)^{3}$
(5) $\left(\frac{2}{5}\right) \times \left(\frac{3}{5}\right)^{2}$

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
If using Method 2 to participate in the lottery until winning one action figure, express the expected value of the number of lottery draws needed using the definition of expected value and the $\sum$ notation, and find its value. (Non-multiple choice question, 4 points)

A store sells a popular action figure through a lottery. Each lottery draw is independent with a probability of winning of $\frac{2}{5}$. Participants can participate in the lottery using one of the following two methods.
Method 1: Pay 225 yuan to get two lottery chances. Stop drawing as soon as you win and receive one action figure. If you fail to win in both draws, you must pay an additional 75 yuan to receive one action figure.
Method 2: Unlimited number of lottery draws, paying 100 yuan per draw.
Assuming there is no limit on spending until obtaining one action figure, find the expected value of the amount paid to obtain one action figure using each of the two lottery methods, and explain the relationship between these two expected values. (Non-multiple choice question, 6 points)

A holiday market stall offers ``test your luck—cute dolls regularly priced at 480 yuan can be purchased for as low as 240 yuan''. The rules are: customers flip a fair coin up to 5 times. If 3 consecutive heads are obtained in the first 3 flips, they can purchase a doll for 240 yuan. If 3 heads are accumulated by the 4th flip, they can purchase for 320 yuan. If 3 heads are accumulated by the 5th flip, they can purchase for 400 yuan. If 3 heads are not accumulated after 5 flips, they can purchase for 480 yuan. The expected value of the amount a customer spends to purchase a doll is (15-1) (15-2) (15-3) yuan.

Let $z _ { n }$ and $w _ { n } ( n = 0,1,2 , \ldots )$ be complex numbers. Consider a bag that contains two red cards and one white card. First, take one card from the bag and return it to the bag. $z _ { k + 1 } ( k = 0,1,2 , \ldots )$ is generated in the following manner based on the color of the card taken.
$$z _ { k + 1 } = \begin{cases} i z _ { k } & \text { if a red card was taken, } \\ - i z _ { k } & \text { if a white card was taken. } \end{cases}$$
Next, take one card from the bag again and return it to the bag. $w _ { k + 1 }$ is generated in the following manner based on the color of the card taken.
$$w _ { k + 1 } = \begin{cases} - i w _ { k } & \text { if a red card was taken, } \\ i w _ { k } & \text { if a white card was taken. } \end{cases}$$
Here, each card is independently taken with equal probability. The initial state is $z _ { 0 } = 1$ and $w _ { 0 } = 1$. Thus, $z _ { n } , w _ { n }$ are the values after repeating the series of the above two operations $n$ times starting from the state of $z _ { 0 } = 1$ and $w _ { 0 } = 1$. Here, $i$ is the imaginary unit.
Answer the following questions.
(1) Show that $\operatorname { Re } \left( z _ { n } \right) = 0$ if $n$ is odd, and that $\operatorname { Im } \left( z _ { n } \right) = 0$ if $n$ is even. Here, $\operatorname { Re } ( z )$ and $\operatorname { Im } ( z )$ represent the real part and the imaginary part of $z$ respectively.
(2) Let $P _ { n }$ be the probability of $z _ { n } = 1$, and $Q _ { n }$ be the probability of $z _ { n } = i$. Find recurrence equations for $P _ { n }$ and $Q _ { n }$.
(3) Find the probabilities of $z _ { n } = 1 , z _ { n } = i , z _ { n } = - 1$, and $z _ { n } = - i$ respectively.
(4) Show that the expected value of $z _ { n }$ is $( i / 3 ) ^ { n }$.
(5) Find the probability of $z _ { n } = w _ { n }$.
(6) Find the expected value of $z _ { n } + w _ { n }$.
(7) Find the expected value of $z _ { n } w _ { n }$.

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 $P \left( X _ { 3 } = 1 \mid X _ { 1 } = 0 \wedge X _ { 2 } = 1 \wedge X _ { 4 } = 1 \right)$.

Consider a game where points are awarded in $n$ independent trials. In each trial, either $+1$ or $-1$ is awarded and both outcomes have the same probability of $1/2$. Let $X _ { k }$ be the point awarded in the $k ^ { \text {th } }$ trial $( 1 \leq k \leq n )$, and $S _ { k } = \sum _ { i = 1 } ^ { k } X _ { i }$. In the following questions, $n$ is an even integer such that $n \geq 4$, and $t$ is an even integer such that $2 \leq t \leq n$.
I. Obtain the probability for $S _ { 4 } = 0$.
II. Let $P _ { n } ( t )$ be the probability for $S _ { n } = t$. Find $P _ { n } ( t )$.
III. Let $P _ { n } ^ { + } ( t )$ be the probability for $S _ { 1 } = 1$ and $S _ { n } = t$. Find $P _ { n } ^ { + } ( t )$.
IV. Let $P _ { n } ^ { - } ( t )$ be the probability for $S _ { 1 } = - 1$ and $S _ { n } = t$. Find $P _ { n } ^ { - } ( t )$.
V. Let $Q _ { n } ( t )$ be the probability that all of the variables $\left\{ S _ { j } \right\}$ $( j = 1,2 , \cdots , n - 1 )$ are greater than zero and $S _ { n } = t$. Express $Q _ { n } ( t )$ with $P _ { n } ^ { + } ( t )$ and $P _ { n } ^ { - } ( t )$. Then, express $Q _ { n } ( t )$ with $P _ { n } ( t )$.
VI. Obtain the probability that all of the variables $\left\{ S _ { j } \right\} ( j = 1,2 , \cdots , n )$ are greater than zero.

Problem 6

Consider an electric vehicle charging station with a single charger installed and let us observe the number of vehicles at the station at regular time intervals.
Arriving vehicles at the station are lined up in the queue in the order of arrival, and only the first vehicle in the queue can be charged. In the interval between one observation and the next observation, assume that one new vehicle arrives with probability $p ( 0 < p < 1 )$, and that the vehicle charging at the head of the queue completes charging with probability $q ( 0 < q < 1 )$. Here, assume that $p$ and $q$ are constants and $p + q < 1$.
The queue can accommodate $N ( N \geq 2 )$ vehicles, including the vehicle being charged at the head of the queue, and the $( N + 1 )$-th vehicle shall give up and leave the station without queuing up. The vehicle which completes charging leaves the station immediately.
In the interval between one observation and the next observation, either only one or no vehicles arrive at the station and either only one or no vehicles complete charging. Moreover, assume that both arrival of new vehicle and completion of charging for the first vehicle do not occur together in any one interval.
I. When there are $i ( 0 < i < N )$ vehicles in the queue, find the probability for the following condition: no new vehicle arrives and the first vehicle does not leave in the interval between one observation and the next observation.
Let $\pi _ { i } ( 0 \leq i \leq N )$ be the probability that $i$ vehicles are in the queue in the steady state.
II. Express the relationship between $\pi _ { i }$ and $\pi _ { i + 1 }$. Here, $i \leq N - 1$.
III. Express $\pi _ { i }$ using $p , q$ and $N$.
IV. Find the expected value of the number of vehicles at the station in the steady state using $p , q$ and $N$. Here, $p < q$.