There are 6 weights of 1 unit, 3 weights of 2 units, and 1 empty bag. Using one die, the following trial is performed. (Here, the unit of weight is g.)
Roll the die once. If the number shown is 2 or less, put one weight of 1 unit into the bag. If the number shown is 3 or more, put one weight of 2 units into the bag.
Repeat this trial until the total weight of the weights in the bag is first greater than or equal to 6. Let $X$ be the random variable representing the number of weights in the bag. The following is the process of finding the probability mass function $\mathrm { P } ( X = x ) ( x = 3,4,5,6 )$ of $X$.
(i) The event $X = 3$ is the case where 3 weights of 2 units are in the bag, so $$\mathrm { P } ( X = 3 ) = \text{ (a) }$$ (ii) The event $X = 4$ can be divided into the case where the total weight of weights put in by the third trial is 4 and a weight of 2 units is put in on the fourth trial, and the case where the total weight of weights put in by the third trial is 5. Therefore, $$\mathrm { P } ( X = 4 ) = \left( \text{ (b) } + { } _ { 3 } \mathrm { C } _ { 1 } \left( \frac { 1 } { 3 } \right) ^ { 1 } \left( \frac { 2 } { 3 } \right) ^ { 2 } \right) \times \frac { 2 } { 3 }$$ (iii) The event $X = 5$ can be divided into the case where the total weight of weights put in by the fourth trial is 4 and a weight of 2 units is put in on the fifth trial, and the case where the total weight of weights put in by the fourth trial is 5. Therefore, $$\mathrm { P } ( X = 5 ) = { } _ { 4 } \mathrm { C } _ { 4 } \left( \frac { 1 } { 3 } \right) ^ { 4 } \left( \frac { 2 } { 3 } \right) ^ { 0 } \times \frac { 2 } { 3 } + \text{ (c) }$$ (iv) The event $X = 6$ is the case where the total weight of weights put in by the fifth trial is 5, so $$\mathrm { P } ( X = 6 ) = \left( \frac { 1 } { 3 } \right) ^ { 5 }$$ If the values corresponding to (a), (b), (c) are $a , b , c$ respectively, what is the value of $\frac { a b } { c }$? [4 points]
(1) $\frac { 4 } { 9 }$
(2) $\frac { 7 } { 9 }$
(3) $\frac { 10 } { 9 }$
(4) $\frac { 13 } { 9 }$
(5) $\frac { 16 } { 9 }$

A bag contains 10 balls labeled with the number 1, 20 balls labeled with the number 2, and 30 balls labeled with the number 3. One ball is randomly drawn from the bag, the number on the ball is confirmed, and then the ball is returned. This procedure is repeated 10 times, and the sum of the 10 numbers confirmed is the random variable $Y$. The following is the process of finding the mean $\mathrm { E } ( Y )$ and variance $\mathrm { V } ( Y )$ of the random variable $Y$.
Let the 60 balls in the bag be the population. When one ball is randomly drawn from this population, let the number on the ball be the random variable $X$. The probability distribution of $X$, that is, the probability distribution of the population, is shown in the following table.

$X$	1	2	3	Total
$\mathrm { P } ( X = x )$	$\frac { 1 } { 6 }$	$\frac { 1 } { 3 }$	$\frac { 1 } { 2 }$	1

Therefore, the population mean $m$ and population variance $\sigma ^ { 2 }$ are $$m = \mathrm { E } ( X ) = \frac { 7 } { 3 } , \quad \sigma ^ { 2 } = \mathrm { V } ( X ) = \text { (가) }$$
When a sample of size 10 is randomly extracted from the population and the sample mean is $\bar { X }$, $$\mathrm { E } ( \bar { X } ) = \frac { 7 } { 3 } , \quad \mathrm {~V} ( \bar { X } ) = \text { (나) }$$
Let the number on the $n$-th ball drawn from the bag be $X _ { n }$. Then $$Y = \sum _ { n = 1 } ^ { 10 } X _ { n } = 10 \bar { X }$$ so $$\mathrm { E } ( Y ) = \frac { 70 } { 3 } , \quad \mathrm {~V} ( Y ) = \text { (다) }$$
When the numbers that fit (가), (나), and (다) are $p , q , r$ respectively, what is the value of $p + q + r$? [4 points]
(1) $\frac { 31 } { 6 }$
(2) $\frac { 11 } { 2 }$
(3) $\frac { 35 } { 6 }$
(4) $\frac { 37 } { 6 }$
(5) $\frac { 13 } { 2 }$

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) $\frac { 13 } { 16 }$

There is a basket containing at least 10 white balls and at least 10 black balls, and an empty bag. Using one die, the following trial is performed.
Roll the die once. If the result is 5 or more, put 2 white balls from the basket into the bag. If the result is 4 or less, put 1 black ball from the basket into the bag.
When the above trial is repeated 5 times, let $a _ { n }$ and $b _ { n }$ denote the number of white balls and black balls in the bag after the $n$-th trial ($1 \leq n \leq 5$) respectively. Given that $a _ { 5 } + b _ { 5 } \geq 7$, what is the probability that there exists a natural number $k$ ($1 \leq k \leq 5$) such that $a _ { k } = b _ { k }$? If this probability is $\frac { q } { p }$, find the value of $p + q$. (Here, $p$ and $q$ are coprime natural numbers.) [4 points]

When 4 coins are tossed simultaneously, let $X$ be the random variable representing the number of coins showing heads. Define the discrete random variable $Y$ as $$Y = \begin{cases} X & (\text{if } X \text{ takes the value } 0 \text{ or } 1) \\ 2 & (\text{if } X \text{ takes a value of } 2 \text{ or more}) \end{cases}$$ Find the value of $\mathrm{E}(Y)$. [3 points]
(1) $\frac{25}{16}$
(2) $\frac{13}{8}$
(3) $\frac{27}{16}$
(4) $\frac{7}{4}$
(5) $\frac{29}{16}$

A discrete random variable $X$ takes values that are integers from 0 to 4, and $$\mathrm { P } ( X = x ) = \left\{ \begin{array} { c l } \frac { | 2 x - 1 | } { 12 } & ( x = 0,1,2,3 ) \\ a & ( x = 4 ) \end{array} \right.$$ What is the value of $\mathrm { V } \left( \frac { 1 } { a } X \right)$? (Here, $a$ is a nonzero constant.) [3 points]
(1) 36
(2) 39
(3) 42
(4) 45
(5) 48

6. The probability distribution of the random variable $\xi$ is given in the table below:

$\mathbf { x }$	7	$\mathbf { 8 }$	$\mathbf { 9 }$	$\mathbf { 1 0 }$
$\mathbf { P } ( \xi = x )$	0.3	0.35	0.2	0.15

Then the expected value of the random variable $\xi$ is $\_\_\_\_$ $8.2$
Analysis: This examines the definition of expected value. $\mathrm { E } \xi = 7 \times 0.3 + 8 \times 0.35 + 9 \times 0.2 + 10 \times 0.15 = 8.2$

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.

19. (12 points) The one-way driving time T between a school's new and old campuses depends only on road conditions. A sample of 100 observations was collected with the following results:

T (minutes)

25

30

35

40

Frequency

20

30

40

10

(I) Find the probability distribution of T and the mathematical expectation ET; (II) Professor Liu drives from the old campus to the new campus for a 50-minute lecture, then immediately returns to the old campus. Find the probability that the total time from leaving the old campus to returning is no more than 120 minutes.

15. Teams A and B are playing a best-of-seven basketball series (the series ends when one team wins four games). Based on previous results, Team A's home and away arrangement is ``home, home, away, away, home, away, home'' in order. The probability that Team A wins at home is 0.6, and the probability that Team A wins away is 0.5. Each game is independent. The probability that Team A wins 4-1 is $\_\_\_\_$.

18. (12 points) In an 11-point table tennis match, each point won scores 1 point. When the score reaches 10:10, players alternate serves, and the first player to score 2 more points wins the match. Two students, A and B, play a singles match. Assume that when A serves, A scores with probability 0.5; when B serves, A scores with probability 0.4. The results of each point are independent. After a certain match reaches 10:10 with A serving first, the two players play $X$ more points before the match ends.
(1) Find $P ( X = 2 )$;
(2) Find the probability of the event ``$X = 4$ and A wins''.

21. Solution: The possible values of $X$ are $-1, 0, 1$.
$$\begin{aligned} & P(X = -1) = (1-\alpha)\beta, \\ & P(X = 0) = \alpha\beta + (1-\alpha)(1-\beta), \\ & P(X = 1) = \alpha(1-\beta), \end{aligned}$$
Therefore the distribution table of $X$ is

$X$	$-1$	$0$	$1$
$P$	$(1-\alpha)\beta$	$\alpha\beta + (1-\alpha)(1-\beta)$	$\alpha(1-\beta)$

(2) (i) From (1) we have $a = 0.4$, $b = 0.5$, $c = 0.1$. Therefore $p_i = 0.4p_{i-1} + 0.5p_i + 0.1p_{i+1}$, so $0.1(p_{i+1} - p_i) = 0.4(p_i - p_{i-1})$, that is $p_{i+1} - p_i = 4(p_i - p_{i-1})$. Since $p_1 - p_0 = p_1 \neq 0$, the sequence $\{p_{i+1} - p_i\}$ $(i = 0, 1, 2, \cdots, 7)$ is a geometric sequence with common ratio 4 and first term $p_1$.
(ii) From (i) we obtain $p_8 = p_8 - p_7 + p_7 - p_6 + \cdots + p_1 - p_0 + p_0 = (p_8 - p_7) + (p_7 - p_6) + \cdots + (p_1 - p_0) = \frac{4^8 - 1}{3}p_1$. Since $p_8 = 1$, we have $p_1 = \frac{3}{4^8 - 1}$, therefore $p_4 = (p_4 - p_3) + (p_3 - p_2) + (p_2 - p_1) + (p_1 - p_0) = \frac{4^4 - 1}{3}p_1 = \frac{1}{257}$. $p_4$ represents the probability of ultimately concluding that drug A is more effective. From the calculation result, we can see that when drug A has a cure rate of 0.5 and drug B has a cure rate of 0.8, the probability of concluding that drug A is more effective is $p_4 = \frac{1}{257} \approx 0.0039$. The probability of reaching an incorrect conclusion is very small, indicating that this experimental scheme is reasonable.

18.
(1)
The probability distribution of $X$ is:
$$\begin{gathered} P ( X = 0 ) = 0.2 \\ P ( X = 20 ) = 0.8 \times ( 1 - 0.6 ) = 0.32 \\ P ( X = 100 ) = 0.8 \times 0.6 = 0.48 \end{gathered}$$
(2)
Given that type A questions are answered first, the mathematical expectation is...

Given that the premium for a certain insurance is 0.4 ten thousand yuan. For the first 3 claims, each claim pays 0.8 ten thousand yuan; the 4th claim pays 0.6 ten thousand yuan.

Number of Claims	0	1	2	3	4
Number of Policies	800	100	60	30	10

A sample of 100 policies is drawn from the population. Using frequency to estimate probability:
(1) Find the probability that a randomly selected policy has at least 2 claims;
(2) (i) Gross profit is the difference between premium and claim amount. Let gross profit be $X$. Estimate the mathematical expectation of $X$;
(ii) If policies with no claims have their premium reduced by 4\% in the next insurance period, and policies with claims have their premium increased by 20\%, estimate the mathematical expectation of gross profit for the next insurance period.

A box contains 5 identical balls labeled $1$ to $5$. Drawing with replacement three times, let $X$ denote the number of distinct balls drawn at least once. Then the mathematical expectation $E(X) = $ $\_\_\_\_$ .

A box contains 5 balls labeled $1$ to $5$. If we draw with replacement three times, let $X$ denote the number of distinct balls drawn at least once. Then $E(X) = $ $\_\_\_\_$ .

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 $m$, $p_{2m+1} - q_{2m+1} < p_{2m} - q_{2m} < p_{2m+2} - q_{2m+2}$.

A green die and a red die are rolled simultaneously. The random variable $X$ describes the sum of the two numbers rolled. Give all values that the random variable $X$ can take and determine the probability $P ( X = 7 )$.

Describe how one can immediately recognize that the expected value of $Y$ is equal to 2.

The variance of $Y$ is equal to $\frac { 11 } { 8 }$. Determine the values of $a$ and $b$.

Justify that $X$ is not binomially distributed.

Determine for the case $n = 5$ the probability that not all three badges have the same motif.

Justify that the probability $P ( X = 4 )$ agrees with the probability $P ( X = 10 )$.

The probability distributions of $X$ and $Y$ are each represented by one of the following diagrams I, II and III. Assign $X$ and $Y$ to the appropriate diagram and justify your assignment.\n[Figure]\n[Figure]\n[Figure]

Give an event in the context of the problem whose probability can be calculated using the term $\left( \frac { 3 } { 4 } \right) ^ { 5 }$, and justify your answer.