在公平的抛硬币中得到正面的概率是多少？
处理随机事件发生的数学分支称为概率。它在数学中用于预测事件发生的可能性。任何事件的概率只能在 0 到 1 之间，也可以写成百分比的形式。
可能性
事件 A 的概率一般写为 P(A)。这里P代表可能性，A代表事件。它说明事件即将发生的可能性。事件的概率只能存在于 0 和 1 之间，其中 0 表示事件不会发生，即不可能，1 表示肯定会发生，即确定性。
如果不确定事件的结果，请借助某些结果的概率，以及它们发生的可能性。为了正确理解概率，我们以抛硬币为例，会有两种可能的结果——正面或反面。
得到正面的概率是一半。众所周知，概率是一半/一半或 50%，因为该事件是同等可能的事件并且是互补的，因此出现正面或反面的可能性为 50%。
概率公式
Probability of an event = Favorable outcomes / Total number of outcomes
P(A) = Favorable outcomes / Total number of outcomes
概率论的一些术语
- 实验：为产生结果而进行的操作或试验称为实验。
- 样本空间：一个实验共同构成了所有可能结果的样本空间。例如，抛硬币的样本空间是正面和反面。
- 有利结果：产生所需结果的事件称为有利结果。例如，如果同时掷出两个骰子，则将两个骰子上的数字之和设为 4 的可能或有利结果是 (1, 3)、(2, 2) 和 (3, 1)。
- 试验：试验意味着进行随机实验。
- 随机实验：随机实验是具有明确定义的结果集的实验。例如，当我们抛硬币时，我们会领先或落后，但结果无法确定哪个会出现。
- 事件：事件是随机实验的结果。
- 同等可能的事件：同等可能的事件是具有相同机会或概率发生的罕见事件。这里一个事件的结果独立于另一个。例如，当抛硬币时，领先或落后的机会均等。
- 穷举事件：穷举事件是当实验的所有结果的集合等于样本空间时。
- 互斥事件：不能同时发生的事件称为互斥事件。例如，气候可以是冷的或热的。一次又一次的天气是无法体验的。
- 互补事件：只有两种结果的可能性，即一个事件是否会发生，例如一个人会吃或不吃食物，买自行车或不买自行车等，都是互补事件的例子。
一些概率公式
- 加法规则：两个事件的并集，比如 A 和 B，那么，
P(A or B) = P(A) + P(B) – P(A∩B)
P(A ∪ B) = P(A) + P(B) – P(A∩B)
- 互补规则：如果一个实验有两个可能的事件，那么一个事件的概率将是另一个事件的补码。例如，如果 A 和 B 是两个可能的事件，那么，
P(B) = 1 – P(A) or P(A’) = 1 – P(A).
P(A) + P(A′) = 1.
- 条件规则：当给定一个事件的概率并且需要第二个事件的概率时，第一个给定，那么 P(B, given A) = P(A and B), P(A, given B)。反之亦然，
P(B∣A) = P(A∩B)/P(A)
- 乘法规则：另外两个事件的交集，即事件 A 和 B 需要同时发生。然后，
P(A and B) = P(A) P(B).
P(A∩B) = P(A) P(B∣A)
在公平的抛硬币中得到正面的概率是多少？
解决方案：
If a fair coin is tossed then the sample space will be {H, T}
Therefore total number of event = 2
Probability of having head = 1
Probability of an event = (number of favorable event) / (total number of event)
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
类似问题
问题1：连续翻转5个正面的机会是多少？
解决方案：
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 2 heads in a row = probability of getting head first time × probability of getting head second time.
Probability of getting 2 head in a row = (1/2) × (1/2).
Therefore, the probability of getting 5 heads in a row = (1/2)5.
问题2：连续翻转20个正面的机会是多少？
解决方案：
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one head = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 heads in a row = probability of getting head first time × probability of getting head second time × probability of getting head third time
Probability of getting 3 head in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 20 heads in a row = (1/2)20
问题3：连续翻转10个尾巴的机会是多少？
解决方案：
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
Here, if Tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 tails in a row = probability of getting tail first time × probability of getting tail second time × probability of getting tail third time
Probability of getting 3 tails in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 10 tails in a row = (1/2)10
问题 4：抛一个公平的硬币时，得到尾巴的概率是多少？
解决方案：
If a fair coin is tossed then the sample space will be {H, T}
Therefore total number of event = 2
Probability of having tail = 1
Probability of an event = (number of favorable event) / (total number of event)
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
问题 5：连续翻转 20 条尾巴的几率是多少？
解决方案：
Probability of an event = (number of favorable event) / (total number of event).
P(B) = (occurrence of Event B) / (total number of event).
Probability of getting one tail = 1/2.
Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed.
Probability of getting 3 tails in a row = probability of getting tail first time × probability of getting tail second time × probability of getting tail third time
Probability of getting 3 tails in a row = (1/2) × (1/2) × (1/2)
Therefore, the probability of getting 20 tails in a row = (1/2)20