抛硬币 20 次得到 20 个正面的概率是多少？
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。
The probability of event A is generally written as P(A).
Here P represents the possibility and A represents the event. It states how likely an event is about to happen. The probability of an event can exist only between 0 and 1 where 0 indicates that event is not going to happen i.e. Impossibility and 1 indicates that it is going to happen for sure i.e. Certainty
如果我们不确定某个事件的结果，我们会借助某些结果的概率——它们发生的可能性有多大。为了正确理解概率，我们举一个抛硬币的例子：
将有两种可能的结果——正面或反面。
得到正面的概率是一半。您可能已经知道概率是一半/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, 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，如果我们同时掷两个骰子，那么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，但我们不确定会出现哪个结果。
- 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，当我们抛硬币时，得到正面或反面的机会均等。
- 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。我们不能一次又一次地经历同样的天气。
- 补充事件：只有两个结果是一个事件的可能性会发生与否。就像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，然后
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：当给定事件的概率并且需要第二个事件的概率时，第一个给定的，那么
P(B, given A) = P(A and B), P(A, given B). It can be vice versa
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)
抛硬币 20 次得到 20 个正面的概率是多少？
解决方案：
20 times coin tosses. This means,
Total observations = 220 (According to binomial concept)
Required outcome → 20 Heads {H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,H,}
This can occur only ONCE!
Thus, required outcome =1
Now put the probability formula
Probability (20 Heads) = (1⁄2)20 = 1⁄1048576
Hence, the probability that it will always land on the HEAD side will be, (1⁄2)20 = 1⁄1048576
类似问题
问题1：连续翻转15次正面的机会是多少？
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 15 heads in a row = (1/2)15
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 x 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 x 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： 10 个硬币中出现 10 个正面的概率是多少？
解决方案：
If the coin is fair, then the probability of getting either a head or a tail is 1/2.
So the probability of getting 10 heads in a row is 1/210