掷硬币 12 次得到 4 个正面的概率是多少？
概率是处理随机事件发生的数学分支。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)
掷硬币 12 次得到 4 个正面的概率是多少？
解决方案：
Use the binomial distribution.
Lets suppose that the number of heads is r that represents the head times and in this case r = 4
Assuming that the coin is unbiased, you have a probability of success ‘p’(where p is considered as success) is 1/2 and the probability of failure ‘q’ is (where q is considered as failure).
The number of trials is represented by the ’n’ and here n = 12.
Now use the function for a binomial distribution:
P(R = r) = nCrprqn-r
P(R = 4) = (12C4)(1/2)4(1/2)12-4
= 495/4096
So the probability of flipping a coin 12 times and getting heads 4 times is 495/4096.
类似问题
问题1：连续翻转10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 10 heads in a row = (1/2)10.
问题2：连续翻转12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)12
问题3：抛硬币20次，出现6次正面的概率是多少？
解决方案：
Use the binomial distribution.
Lets suppose that the number of heads is r that represents the head times and in this case r = 6
Assuming that the coin is unbiased, you have a probability of success ‘p’(where p is considered as success) is 1/2 and the probability of failure ‘q’ is (where q is considered as failure).
The number of trials is represented by the ’n’ and here n = 20.
Now use the function for a binomial distribution:
P(R = r) = nCr pr qn-r
P(R = 6) = (20C6)(1/2)6(1/2)12-6
= (20C6) (1/2)20
= 38760/1048576
= .0369644
So the probability of flipping a coin 20 times and getting heads 6 times is 0.0369644