如何找到抛硬币的概率？

概率是处理特定事件发生的数学分支。它基本上指定了某事发生的可能性。例子：晴天下雨的概率是多少？概率的范围在 0 和 1 之间。不可能事件的概率为 0，确定事件的概率为 1。概率也可以用百分比表示。

与概率相关的术语

• 实验：任何提供明确定义结果的操作都称为实验。例如：掷硬币或掷骰子是一种实验。
• 随机实验：在任何实验中，所有可能的结果，但不知道会发生哪个确切结果。这称为随机实验。例如：通过抛硬币，获得正面或反面，但不确定只会出现正面或尾部。
• 样本空间：样本空间是所有可能结果的集合。示例：投掷硬币时，我们有 2 个结果：正面和反面。
• 试验：这是一个进行试验并记录结果的过程。例如：从一副 52 张牌中取出一张牌。
• 事件：实验的每个结果称为事件。例如：投掷硬币获得正面是一个事件。
• 独立事件：当一个事件的发生不受另一事件发生的影响时，称为独立事件。例如，一个人可以同时掷硬币和掷骰子，因为它们是不同的事件。
• 穷举事件：如果两个事件的并集等于样本空间，则称这两个事件是穷举的。
• 排他事件：当两个事件不能同时发生或两个事件不相交时，称为排他事件。例如：在抛硬币时，一个人可以得到正面或反面，但不能两者兼而有之。

概率公式

The formula for probability is,

Probability = Number of favorable outcomes/Total number of outcomes

For example:

P(A) = Number of ways A occur/Total number of outcomes

如何找到抛硬币的概率？

It is known that a coin has two sides: Heads and Tails. It is not known which outcome will occur but one knows that there are 2 chances: one is head and the other is tail. It is a random experiment. Suppose there is an unbiased coin. So the total number of outcomes = 2 (Since 1 head and 1 tail) and one wants the head to occur. What is the probability of occurrence of head? Since 1 head is present in 1 coin and the total number of outcomes is 2

The probability of occurrence of head = 1/2

and the probability of occurrence of tail = 1/2

Finding probability for multiple coins

The probability of 1 coin is found but what about 2 coins or more than that. Let’s check it out. For 2 coins there are four net outcomes {HH, TH, HT, TT} since on first coin head or tail can occur. Similar is for second coin. So from the above case one wants to know the probability of finding,

• 2 heads
• 1 tail.

For first case,

Favourable outcomes = {HH} = 1

Probability of 2 heads is = 1/4

For second case,

Favorable outcomes = {TH, HT} = 2 (Here it is specifically mentioned that we have to find probability of 1 tail so 2 tails is not considered}

Probability of 1 tail = 2/4 = 1/2

But as the number of coins increases outcomes also increased so it’s not possible to find the favorable outcomes. Since tossing coins is independent event we use binomial distribution. The formula for binomial distribution is,

P(X) = ⁿC_x  × p^x × (1 – p)^{n – x}

Where n is total number of trials

x is a favorable trial, p is the probability of the favourable outcome. 

1 – p is the probability of unfavorable outcome. 

Let’s find the probability of 1 tail using 2 coins 

Probability of occurrence of tail = 0.5

Total number of trials = 2

Probability of head = 0.5

So the probability is = ²C₁ × (0.5) ¹ ×  (1 – 0.5) ¹ =0.5

Let us find the probability of 1 tail using 3 coins. Total trials is 3, Favorable trial is 1

Probability = ³C₁ × (0.5) ¹ × (1 – 0.5) ^{3 – 1} = {3/(1! 2! )} × 0.5 × 0.5² = 3 × (1/2) × (1/2) × (1/2) = 3/8 which is our answer. So the tossing of coins is not at all tough. Just remembering the formula and understanding the concept will help. 

示例问题

问题1：投掷硬币3次。每枚硬币出现同一张脸的概率是多少？ （提示：有利事件为 HHH、TTT）

解决方案：

The sample space are HTH, HHT, HHH, HTT, TTH, THT, TTT, THT

Total number of sample space events = 8

Favourable events = HHH, TTT

Number of favourable events = 2

Probability of getting same face on three coins = 2/8 = 1/4 = 0.25

问题2：有10个硬币，同时翻转。找出得到 5 个正面的概率。

解决方案：

Probability of getting head on 1 coin (p) = 0.50

Probability of getting tail on 1 coin (q) = 0.5

Number of coins (n) = 10

Number of heads(r) = 5

Using the formula of Binomial Distribution,

ⁿC_rp^rq^{n – r}

¹⁰C₅ (0.5) ⁵ (0.5)^{10 – 5} = [(10!) /(5! × 5!)] × (0.5)¹⁰ = 63/256 = 0.2461

问题3：同时抛两枚硬币。得到头和尾的概率是多少？

解决方案：

The sample space are TT, HH, TH, HT

Number of sample spaces = 2

Favourable events = HT, TH

Number of favourable events = 2

Probability of getting 1 head and 1 tail = 2/4 = 1/2 = 0.5