概率基础练习题 - 掌握概率计算与样本空间分析
以下是6道综合练习题，涵盖概率基础概念、样本空间分析和分组数据概率估算等核心内容。
Two coins are tossed. Find the probability of both coins showing the same outcome.
解答过程：
步骤1：确定样本空间
投掷两枚硬币的样本空间：\( \{HH, HT, TH, TT\} \)，共4种等可能结果
步骤2：确定目标事件
两枚硬币显示相同结果的事件：\( \{HH, TT\} \)，共2种结果
步骤3：计算概率
\( P(\text{相同结果}) = \frac{2}{4} = \frac{1}{2} = 0.5 \)
Two six-sided dice are rolled and their product, \( X \), is recorded.
a) Draw a sample space diagram showing all the possible outcomes of this experiment.
b) Find the probability of each event:
i) \( X = 24 \)
ii) \( X < 5 \)
iii) \( X \) is even.
解答过程：
a) 样本空间图
两个骰子的乘积样本空间包含36种结果，其中乘积为24的结果只有\( (4,6) \)和\( (6,4) \)
b) 概率计算
i) \( X = 24 \)：只有\( (4,6) \)和\( (6,4) \)两种结果
\( P(X = 24) = \frac{2}{36} = \frac{1}{18} \)
ii) \( X < 5 \)：乘积小于5的结果有\( (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (3,1) \)
\( P(X < 5) = \frac{7}{36} \)
iii) \( X \) is even：乘积为偶数的结果包括所有至少有一个偶数的组合
\( P(X \text{ is even}) = \frac{27}{36} = \frac{3}{4} \)
The masses of 140 adult Bullmastiff dogs are recorded in a table. One dog is chosen at random.
| Mass, \( m \) (kg) | \( 45 \leq m < 48 \) | \( 48 \leq m < 51 \) | \( 51 \leq m < 54 \) | \( 54 \leq m < 57 \) | \( 57 \leq m < 60 \) | \( 60 \leq m < 63 \) |
|---|---|---|---|---|---|---|
| Frequency | 17 | 25 | 42 | 33 | 21 | 2 |
a) Find the probability that the dog has a mass of 54kg or more.
b) Find the probability that the dog has a mass between 48kg and 57kg.
c) Is it more or less likely that a Bullmastiff chosen at random has a mass under 53kg? State one assumption that you have made in making your decision.
解答过程：
a) 54kg或以上的概率
54kg或以上的狗数：\( 33 + 21 + 2 = 56 \)
\( P(\text{质量} \geq 54\text{kg}) = \frac{56}{140} = \frac{2}{5} = 0.4 \)
b) 48kg到57kg之间的概率
48kg到57kg之间的狗数：\( 25 + 42 + 33 = 100 \)
\( P(48\text{kg} < \text{质量} < 57\text{kg}) = \frac{100}{140} = \frac{5}{7} \)
c) 53kg以下的概率
53kg以下的狗数：\( 17 + 25 + 42 \times \frac{2}{3} = 17 + 25 + 28 = 70 \)
\( P(\text{质量} < 53\text{kg}) = \frac{70}{140} = \frac{1}{2} = 0.5 \)
因此，随机选择的狗质量在53kg以下的概率为0.5，与53kg或以上的概率相等。
假设：数据在组内均匀分布。
The lengths, in cm, of 240 koalas are recorded in a table. One koala is chosen at random.
| Length, \( l \) (cm) | \( 65 \leq l < 70 \) | \( 70 \leq l < 75 \) | \( 75 \leq l < 80 \) | \( 80 \leq l < 85 \) | \( 85 \leq l < 90 \) |
|---|---|---|---|---|---|
| Frequency (male) | 20 | 14 | 24 | 47 | 31 |
| Frequency (female) | 4 | 15 | 32 | 27 | 26 |
a) Find the probability that the koala is female.
b) Find the probability that the koala is less than 80cm long.
c) Find the probability that the koala is a male between 75cm and 85cm long.
d) Estimate the probability that a koala chosen at random is juvenile (under 72cm long). State one assumption you have made in making your estimate.
解答过程：
步骤1：计算总数
雄性总数：\( 20 + 14 + 24 + 47 + 31 = 136 \)
雌性总数：\( 4 + 15 + 32 + 27 + 26 = 104 \)
总数量：\( 136 + 104 = 240 \)
a) 雌性概率
\( P(\text{雌性}) = \frac{104}{240} = \frac{13}{30} \)
b) 长度小于80cm的概率
小于80cm的总数：\( (20+4) + (14+15) + (24+32) = 24 + 29 + 56 = 109 \)
\( P(\text{长度} < 80\text{cm}) = \frac{109}{240} \)
c) 雄性且75-85cm长的概率
符合条件的雄性数：\( 24 + 47 = 71 \)
\( P(\text{雄性且75-85cm}) = \frac{71}{240} \)
d) 幼体（小于72cm）概率估算
65-70cm组：\( 24 \times \frac{5}{5} = 24 \)（全部）
70-75cm组：\( 29 \times \frac{2}{5} = 11.6 \)（估算）
幼体总数：\( 24 + 11.6 = 35.6 \)
\( P(\text{幼体}) = \frac{35.6}{240} = \frac{89}{600} \)
假设：数据在组内均匀分布。
The histogram shows the distribution of masses, in kg, of 70 adult cats.
a) Find the probability that a cat chosen at random has a mass more than 5kg.
b) Estimate the probability that a cat chosen at random has a mass less than 6.5kg.
解答过程：
a) 质量大于5kg的概率
假设直方图显示：
质量大于5kg的猫数：\( 20 + 15 + 10 = 45 \)
\( P(\text{质量} > 5\text{kg}) = \frac{45}{70} = \frac{9}{14} \)
b) 质量小于6.5kg的概率估算
小于6kg的猫数：\( 10 + 15 + 20 = 45 \)
6-7kg组中质量小于6.5kg的猫数：\( 15 \times \frac{0.5}{1} = 7.5 \)
总估算数：\( 45 + 7.5 = 52.5 \)
\( P(\text{质量} < 6.5\text{kg}) = \frac{52.5}{70} = \frac{3}{4} = 0.75 \)
假设：数据在组内均匀分布。
Samira picks one card at random from group \( A \) and one card at random from group \( B \). She records the product, \( Y \), of the two cards as the result of her experiment. Given that \( x \) is an integer and that \( P(Y \text{ is even}) = P(Y \geq 20) \), find the possible values of \( x \).
| Group \( A \) | 2 | 7 | 5 |
|---|---|---|---|
| Group \( B \) | 4 | \( x \) |
解答过程：
步骤1：分析样本空间
从组A选择2、7、5，从组B选择4、\( x \)，共6种组合：
步骤2：分析偶数概率
\( Y \)为偶数的情况：
步骤3：分析\( Y \geq 20 \)的概率
\( Y \geq 20 \)的情况：\( Y = 28, Y = 20, Y = 7x \)（当\( x \geq 3 \)时），\( Y = 5x \)（当\( x \geq 4 \)时）
步骤4：建立等式
当\( x \)为奇数时：\( P(\text{偶数}) = \frac{3}{6} = \frac{1}{2} \)
当\( x \)为偶数时：\( P(\text{偶数}) = \frac{6}{6} = 1 \)
通过分析，当\( x = 3 \)时，\( P(Y \geq 20) = \frac{3}{6} = \frac{1}{2} \)，满足条件。
当\( x = 4 \)时，\( P(Y \geq 20) = \frac{4}{6} = \frac{2}{3} \)，不满足条件。