抛 4 个硬币的概率样本空间是多少？

概率也被称为机会数学。这意味着可能性，即处理可能发生的事件。该值从零到一。在数学中，概率已明确用于估计事情发生的可能性。基本上，概率是预期某事发生的程度。

可能性

为了更准确地理解概率，让我们理解一个掷骰子的例子，可能的结果是——1、2、3、4、5和6。发生任何可能的事情的概率是1/6。由于发生任何事情的可能性相同，因此发生任何有利事情的可能性相同，在这种情况下，它是两个 1/6 或 50/3。

概率公式

P(A) = {Number of favourable affair to A} ⁄ {Total number of affair}

与概率相关的术语

• 实验：任何给出明确结果的功能都称为实验。例如：掷硬币或掷骰子是一种实验。
• 随机实验：在任何实验中，都有可能发生的事情，但不知道会发生哪件事情。这称为随机实验。例如：通过掷硬币，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事件的组。示例：在掷硬币时，我们有 2 个结果：正面和反面。
• 试验：这是一个进行试验并获得好评的过程。例如：从一副 52 张牌中选择一张牌。
• 事件：实验的每个结果称为事件或事件。例如：在掷硬币时得到尾巴是一件事情或事件。
• 独立事件：当一件事的发生不受另一件事的发生影响时，称为独立事件。例如，一个人可以同时掷硬币和掷骰子，因为它们是不相关的事务。
• 穷举事件：如果两个事件或事务的连接等于样本空间，则称它们是穷举的。
• 排他事件：当两个事务不能同时发生或两个事务不相交时，称为排他事件。例如：在翻转硬币时，一个人可以得到正面或反面，但不能两者兼而有之。

抛 4 个硬币的概率样本空间是多少？

解决方案：

Each coin flip has 2 likely events, so the flipping of 4 coins has 2×2×2×2 = 16 likely events. We can summarize all likely events as follows, where H shows a head, and T a tail:

HHHH      THHH

HHHT      THHT

HHTH      THTH

HHTT      THTT

HTHH      TTHH

HTHT      TTHT

HTTH      TTTH

HTTT       TTTT

If we suppose that each single coin is equally probable to come up heads or tails, then each of the above 16 result to 4 flips is equally probable. Each happens a fraction one out of 16 times, or each has a possibility of 1/16.

On the other hand, we could assert that the 1st coin has possibility 1/2 to come up heads or tails, the 2nd coin has possibility 1/2 to come up heads or tails, and so on for the 3rd and 4th coins, so that the possibility for any one particular order of heads and tails is just (1/2)×(1/2)×(1/2)×(1/2)=(1/16).

Now lets ask: what is the possibility that in 4 flips, one gets N heads, where N = 0, 1, 2, 3, or 4. We can get this just by summarizing the number of result above which have the desired figures of heads, and dividing by the total number of likely event

N	#outcomes with N heads	probability to get N heads
0	1	1/16 = 0.0625
1	4	4/16 = 0.25
3	6	6/16 = 0.375
4	4	4/16 = 0.25
5	1	1/16 = 0,0625

             

类似问题

问题1：如果抛四枚硬币，既不是4个正面也不是4个反面的概率是多少？

解决方案：

There can be 16 different probability when 4 coins are tossed:

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT

THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT

There are 14 chances when we have neither 4 Heads nor 4 Tails.

Hence, the possibility or probability of occurring neither 4 Heads nor 4 Tails = 14/16 = 7/8

问题2：如果抛四枚硬币，找出应该有两个正面和两个反面的可能性。

解决方案：

P(A) = {Number of favourable affair to A } ⁄ {Total number of affair}

By tossing four coins, the possibility are (H,H,H,H), (T,H,H,H), (H,T,H,H), (H,H,T,H), (H,H,H,T), (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T), (H,H,T,T), (T,T,T,H), (T,T,H,T), (T,H,T,T), (H,T,T,T), (T,T,T,T) where H shows happening of head while tossing a coin and T shows happening of tail while tossing a coin.

Therefore, Total number of likely affair = 16

Here, the favourable affair is occurring two heads and two tails on tossing four coins.

Clearly, the favourable affair after tossing four coins are (T,T,H,H), (T,H,T,H), (T,H,H,T), (H,T,T,H), (H,T,H,T) and (H,H,T,T).

Therefore, Number of favourable affair = 6

Probability of occurring two heads and two tails =6/16=3/8.

Hence, the possibility that there should be two heads and two tails after tossing four coins is 3/8.

问题3：如果你掷硬币4次，得到所有正面的概率是多少？

解决方案：

Sample space: {(HHHH),(HHHT),(HHTH),(HHTT),(HTHH),(HTHT),(HTTH),(HTTT), (THHH),(THHT),(THTH),(THTT),(TTHH),(TTHT),(TTTH),(TTTT)}

Total number of affairs = 16  

Possibility getting all heads :

P(A) = {Number of favourable affair to A} ⁄ {Total number of affair}

= 1/16  i.e., HHHH