离散随机变量 - 练习题
判断以下变量是否为离散随机变量，并说明理由：
Write down whether or not each of the following is a discrete random variable. Give a reason for each answer.
a) 一组学生中身高的平均值
a) The average height of a group of students
b) 掷一枚硬币直到出现正面所需的次数
b) The number of times a coin is tossed before a head appears
c) 一年的月数
c) The number of months in a year
掷四次骰子，记录6出现的次数Y。写出Y的样本空间。
A fair dice is rolled four times and the number of times it falls with a 6 on the top, Y, is noted. Write down the sample space of Y.
一个袋子包含两个标有数字2的圆盘和两个标有数字3的圆盘。从袋子中随机抽取一个圆盘，记录数字后放回。然后再次抽取一个圆盘并记录数字。
A bag contains two discs with the number 2 on them and two discs with the number 3 on them. A disc is drawn at random from the bag and the number noted. The disc is returned to the bag. A second disc is then drawn from the bag and the number noted.
a) 写出这个实验的所有可能结果。
a) Write down all the possible outcomes of this experiment.
离散随机变量X定义为两个数字的和。
The discrete random variable X is defined as the sum of the two numbers.
b) 以表格和概率函数的形式写出X的概率分布。
b) Write down the probability distribution of X as: i) a table ii) a probability function.
一个偏斜的四面骰子，标有数字1、2、3、4。骰子底面上的数字被建模为随机变量X。
A biased four-sided dice with faces numbered 1, 2, 3 and 4 is rolled. The number on the bottommost face is modelled as a random variable X.
a) 给定P(X = x) = k/x，求k的值。
a) Given that P(X = x) = k/x, find the value of k.
b) 以表格形式写出X的概率分布。
b) Write the probability distribution of X in table form.
c) 求以下概率：
c) Find the probability that:
i) X > 2
ii) 1 < X < 4
iii) X > 4
一个公平的旋转器被旋转，直到它停在红色上，或已经被旋转了四次。随机变量S表示旋转的次数。
A fair spinner is spun until it lands on red or has been spun four times in total. Find the probability distribution of the random variable S, the number of times the spinner is spun.
掷三枚公平的硬币。随机变量X定义为出现的正面数。
Three fair coins are tossed. A random variable X is defined as the number of heads that appear when the three coins are tossed.
a) 写出实验的所有可能结果。
a) Write down all the possible outcomes when the three coins are tossed.
b) 以表格和概率函数的形式写出X的概率分布。
b) Write the probability distribution of X as: i) a table ii) a probability function.
a) 不是离散随机变量，因为身高是在连续尺度上测量的。
a) Is not a discrete random variable as height is measured on a continuous scale.
b) 是离散随机变量，因为它是实验结果的计数。
b) Is a discrete random variable as it is a number that is the result of an experiment.
c) 不是离散随机变量，因为它不随实验变化且不是实验的结果。
c) Is not a discrete random variable as it does not vary and is not the result of an experiment.
Y的样本空间是 {0, 1, 2, 3, 4}，因为掷四次骰子，6出现的次数可能为0、1、2、3或4。
The sample space of Y is {0, 1, 2, 3, 4}, because when rolling a die four times, the number of times 6 appears can be 0, 1, 2, 3, or 4.
a) 所有可能结果：(2,2), (2,3), (3,2), (3,3)
a) All possible outcomes: (2,2), (2,3), (3,2), (3,3)
b) i) 概率分布表格：
b) i) Probability distribution table:
| x | 4 | 5 |
|---|---|---|
| P(X = x) | 1/4 | 3/4 |
ii) 概率函数：P(X = 4) = 1/4, P(X = 5) = 3/4
ii) Probability function: P(X = 4) = 1/4, P(X = 5) = 3/4
a) 概率分布：
a) The probability distribution will be:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | k/1 | k/2 | k/3 | k/4 |
由于这是一个概率分布：\(\sum P(X = x) = 1\)
Since this is a probability distribution: \(\sum P(X = x) = 1\)
\(k(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = 1\)
\(k(\frac{12 + 6 + 4 + 3}{12}) = 1\)
\(k = \frac{12}{25}\)
b) 概率分布：
b) The probability distribution is:
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X = x) | 12/25 | 6/25 | 4/25 | 3/25 |
c) i) P(X > 2) = P(X=3) + P(X=4) = 4/25 + 3/25 = 7/25
c) i) P(X > 2) = P(X=3) + P(X=4) = 4/25 + 3/25 = 7/25
ii) P(1 < X < 4) = P(X=2) + P(X=3) = 6/25 + 4/25 = 10/25 = 2/5
ii) P(1 < X < 4) = P(X=2) + P(X=3) = 6/25 + 4/25 = 10/25 = 2/5
iii) P(X > 4) = 0，因为X的最大值是4。
iii) P(X > 4) = 0 because the maximum value of X is 4.
P(S = 1) 是旋转器第一次就停在红色的概率：\(\frac{2}{5}\)
P(S = 1) is the probability that the spinner lands on red the first time: \(\frac{2}{5}\)
如果旋转器第二次停在红色，它第一次必须停在蓝色：\(\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}\)
If the spinner lands on red on the second spin it must have landed on blue on the first spin: \(\frac{3}{5} \times \frac{2}{5} = \frac{6}{25}\)
类似地，第三次停在红色：\(\frac{3}{5} \times \frac{3}{5} \times \frac{2}{5} = \frac{18}{125}\)
Likewise for landing on red on the third spin: \(\frac{3}{5} \times \frac{3}{5} \times \frac{2}{5} = \frac{18}{125}\)
实验在4次旋转后停止：\(P(S = 4) = 1 - (\frac{2}{5} + \frac{6}{25} + \frac{18}{125}) = \frac{27}{125}\)
The experiment stops after 4 spins: \(P(S = 4) = 1 - (\frac{2}{5} + \frac{6}{25} + \frac{18}{125}) = \frac{27}{125}\)
概率分布表格：
Probability distribution table:
| s | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(S = s) | \(\frac{2}{5}\) | \(\frac{6}{25}\) | \(\frac{18}{125}\) | \(\frac{27}{125}\) |
a) 所有可能结果：HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
a) All possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
b) i) 概率分布表格：
b) i) Probability distribution table:
| x | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P(X = x) | 1/8 | 3/8 | 3/8 | 1/8 |
ii) 概率函数：X的取值：x = 1, 2
ii) Probability function: X can take values: x = 1, 2
| x | p(x) |
|---|---|
| 0 | 0.4 |
| 1 | 0.3 |
| 2 | 0.2 |
| 3 | 0.1 |
| 总和 | 1.0 |
验证：所有概率都是非负的，且总和等于1，满足概率质量函数的性质。
Verification: All probabilities are non-negative, and their sum equals 1, satisfying the properties of a probability mass function.
掷两枚公平的骰子，所有可能的点数组合数为36种。X的概率质量函数如下：
When rolling two fair dice, there are 36 possible combinations. The probability mass function of X is as follows:
| x | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| p(x) | 1/36 | 2/36 | 3/36 | 4/36 | 5/36 | 6/36 | 5/36 | 4/36 | 3/36 | 2/36 | 1/36 |
验证规范性：\(\sum p(x) = \frac{36}{36} = 1\)。
Verify normalization: \(\sum p(x) = \frac{36}{36} = 1\).
这是一个二项分布问题，n=3，p=0.7。概率质量函数为：
This is a binomial distribution problem with n=3, p=0.7. The probability mass function is:
\[p(x) = \binom{3}{x} (0.7)^x (0.3)^{3-x}, \quad x = 0, 1, 2, 3\]
具体计算：
Specific calculations:
\(p(0) = \binom{3}{0} (0.7)^0 (0.3)^3 = 1 \times 1 \times 0.027 = 0.027\)
\(p(1) = \binom{3}{1} (0.7)^1 (0.3)^2 = 3 \times 0.7 \times 0.09 = 0.189\)
\(p(2) = \binom{3}{2} (0.7)^2 (0.3)^1 = 3 \times 0.49 \times 0.3 = 0.441\)
\(p(3) = \binom{3}{3} (0.7)^3 (0.3)^0 = 1 \times 0.343 \times 1 = 0.343\)
验证：0.027 + 0.189 + 0.441 + 0.343 = 1.000
Verification: 0.027 + 0.189 + 0.441 + 0.343 = 1.000
（1）验证概率质量函数性质：
(1) Verify probability mass function properties:
所有概率都是非负的：0.1, 0.3, 0.4, 0.15, 0.05 都 ≥ 0。
All probabilities are non-negative: 0.1, 0.3, 0.4, 0.15, 0.05 are all ≥ 0.
总和等于1：0.1 + 0.3 + 0.4 + 0.15 + 0.05 = 1.0。
Sum equals 1: 0.1 + 0.3 + 0.4 + 0.15 + 0.05 = 1.0.
因此，该分布满足概率质量函数的性质。
Therefore, this distribution satisfies the properties of a probability mass function.
（2）P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.3 + 0.4 = 0.8
(2) P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.3 + 0.4 = 0.8
（3）P(X > 1) = P(X=2) + P(X=3) + P(X=4) = 0.4 + 0.15 + 0.05 = 0.6
(3) P(X > 1) = P(X=2) + P(X=3) + P(X=4) = 0.4 + 0.15 + 0.05 = 0.6
（1）该分布是有限离散型，因为事故次数只可能取0,1,2,3,4,5这有限个值。
(1) This distribution is finite discrete because the number of accidents can only take the finite values 0, 1, 2, 3, 4, 5.
（2）P(X=2或X=3) = P(X=2) + P(X=3) = 0.25 + 0.15 = 0.40
(2) P(X=2 or X=3) = P(X=2) + P(X=3) = 0.25 + 0.15 = 0.40
（3）P(X ≥ 1) = 1 - P(X=0) = 1 - 0.2 = 0.8
(3) P(X ≥ 1) = 1 - P(X=0) = 1 - 0.2 = 0.8
（1）验证非负性和规范性：
(1) Verify non-negativity and normalization:
非负性：对于\(x = 1, 2, 3, \dots\)，\(\frac{1}{2^x} > 0\)，显然满足。
Non-negativity: For \(x = 1, 2, 3, \dots\), \(\frac{1}{2^x} > 0\), obviously satisfied.
规范性：\(\sum_{x=1}^\infty \frac{1}{2^x} = \frac{1/2}{1-1/2} = 1\)（等比数列求和）。
Normalization: \(\sum_{x=1}^\infty \frac{1}{2^x} = \frac{1/2}{1-1/2} = 1\) (geometric series sum).
（2）该随机变量是无限离散型，因为可能取值有无限多个（1,2,3,...）。
(2) This random variable is infinite discrete because it can take infinitely many values (1, 2, 3, ...).
这是一个几何分布问题，每次掷骰子获胜的概率p = 6/36 = 1/6。
This is a geometric distribution problem, with success probability p = 6/36 = 1/6 each time.
概率质量函数为：\(p(x) = (1-p)^{x-1} p = (5/6)^{x-1} (1/6)\)，对于 \(x = 1, 2, 3, \dots\)
The probability mass function is: \(p(x) = (1-p)^{x-1} p = (5/6)^{x-1} (1/6)\), for \(x = 1, 2, 3, \dots\)
例如：
For example:
p(1) = 1/6 ≈ 0.1667
p(2) = (5/6)(1/6) ≈ 0.1389
p(3) = (5/6)^2 (1/6) ≈ 0.1157
p(4) = (5/6)^3 (1/6) ≈ 0.0964