如何求掷多个硬币的概率？

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，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

概率中有一些不常用的术语，例如实验，样本空间，有利结果，试验，随机实验等术语。让我们详细看一下它们的定义，

• 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，当抛硬币时，会得到正面或反面，但结果不确定哪个会出现。
• 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，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。一个人不能一次又一次地经历同样的天气。
• 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。那么 P(A 和 B) = P(A)⋅P(B)。

P(A∩B) = P(A)⋅P(B∣A)

如何求掷多个硬币的概率？

解决方案：

Coin flip probabilities only deal with events related to a single or multiple flips of a  fair coin. A toss of fair coin has an equally likely chance of coming up Heads or Tails.

Sample Space: An experiment together constitutes a sample space for all the possible outcomes. For example, the sample space of tossing a coin is head and tail. For example, whenever a coin is flipped, either heads (H) or tails (T) is obtained. So, here the sample space is {H, T}. Every subset of a sample space is called an event.

P(A) = Favorable outcomes / Total number of outcomes

示例问题

问题1：抛一枚硬币的概率是多少？

解决方案：

To calculate the probability of the event, It contains only one element and sample space contains two elements, so the sample space will be {H, T} 

So, total number of outcome is 2.

What is the probability of a coin landing on tails or head?

The probability of landing on head  is given as: P(A) = Favorable outcomes / Total number of outcomes

= 1/2

Same for the probability of landing on tail, P(A) = Favorable outcomes / Total number of outcomes

= 1/2    

问题2：抛两个硬币的概率是多少？

解决方案：

To calculate the probability of event, by flipping of two coins,

Then the sample space will be {HH, HT, TH, TT}

Total number of outcome = 4

Example: Find the probability of,

• At least two Heads.
• Atmost one Heads and on tail.
• One Tail

P(A) = Favorable outcomes / Total number of outcomes

• Probability of At least two Heads

Favorable outcomes of having two head = HH

= 1

Probability of having two Heads P(A) = Favorable outcomes / Total number of outcomes

= 1/4

• Probability of Atleast one tail and one head

Favorable outcomes of having one tail and one head = HT, TH

= 2

Therefore, 

Probability of At least one Tail and one head P(A) = Favorable outcomes / Total number of outcomes

= 2/4

= 1/2

• Probability of getting one tail

Favorable outcomes of having one tail = HT, TH 

= 2

Total number of outcome = 4

Therefore, probability of having two Tail P(A) = Favorable outcomes / Total number of outcomes

= 2/4

= 1/2

问题3：抛三个硬币的概率是多少？找出这些事件的概率？

• 获得两个头。
• 获得所有尾巴。

解决方案：

To calculate the probability of event, by flipping of three coins

Then the sample space will be {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Total number of outcome = 8

P(A) = Favorable outcomes / Total number of outcomes

So now, 

• Probability of getting two heads 

Sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Here the favourable outcome of having Two heads = 3

So the probability of that event, 

P(A) = Favorable outcomes / Total number of outcomes

= 3/8

• Probability of getting two Tails

Sample space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Here the favourable outcome of having all tails = 3

So the probability of that event,

P(A) = Favorable outcomes / Total number of outcomes

= 3/8

问题 4：一枚硬币被抛 20 次。得到至少 1 条尾巴的概率是多少？

解决方案：

Use the binomial distribution,  

Tossing a coin can give 2 outcomes.

So, tossing a coin 20 times can give (2²⁰) outcomes.

If the outcomes of getting at least one tail are excluded; we will be left with the one and only option of getting all ‘heads’.

So, the probability of getting at least one tail = [{(2²⁰) – 1}/(2²⁰)]

= [1 – {1 / (2²⁰)}].

= 0.999999

问题 5：连续翻转 10 次尾巴的几率是多少？

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 a tails = 1/2.

Tossing a coin is an independent event, it is not dependent on how many times it’s been tossed.

Probability of getting 2 tails in a row = probability of getting tail first time × probability of getting tail second time.

Probability of getting 2 tails in a row  = (1/2) × (1/2).

Similarly, the probability of getting 10 tails in a row = (1/2)¹².