一个骰子连续掷出 3 次 1 的概率是多少？

概率是处理随机事件发生的数学分支。概率在数学中用于预测事件发生的可能性。事件的概率仅在 0 和 1 之间，也可以用百分比表示。

事件 B 的概率通常写为 P(B)。这里P代表可能性，B代表事件。

每当我们不确定某个事件的结果时，我们都会利用某些结果的概率——它们发生的可能性有多大。为了理解概率，我们举一个抛硬币的例子：

有两种可能的结果——正面或反面。

得到正面的概率是一半。您可能已经知道可能性是一半/一半或 50%，因为事件是同样可能的事件并且是互补的，因此在这种情况下出现正面或反面的可能性相同，即 50%。

概率公式

$Probability \space of\space an\space event = \frac{Number\space of\space ways\space it\space can\space occur}{Total\space number\space of\space outcomes}$

$P(A) = \frac{Number\space of\space ways\space A\space occurs}{Total\space number\space of\space outcomes}$

同样可能的事件

掷骰子得到任何数字的概率是 1/6。由于该事件是同样可能的事件，因此在这种情况下获得任何数字的可能性是相同的，它是公平掷骰子的 1/6。

补充活动

只有两种结果的可能性，即一个事件会发生与否。就像一个人会跑或不会跑比赛，买车或不买车等都是互补事件的例子。

一个骰子连续掷出 3 次 1 的几率是多少？

解决方案：

Probability of an event = (number of favorable event) / (total number of event).

P(B) = (Event B) / (total number of event).

Probability of getting 1 = 1/6.

Rolling dice is an independent event, it is not dependent on how many times it’s been rolled.

Probability of getting 1 three times in a row = probability of getting 1 first time × probability of getting 1 second time × probability of getting 1 third time.

Probability of getting 1 three times in a row  = (1/6) × (1/6) × (1/6) = 1/216.

Hence, the probability of getting 1 three times in a row is 0.463%.

类似问题

问题 1. 连续 3 次掷出 2 的几率是多少？

解决方案：

Probability of an event = (number of favorable event) / (total number of event).

P(B) = (Event B) / (total number of event).

Probability of getting 2 = 1/6.

Rolling dice is an independent event, it is not dependent on how many times it’s been rolled.

Probability of getting 2 three times in a row = probability of getting 2 first time×probability of getting 2 second time×probability of getting 2 third time.

Probability of getting 2 three times in a row  = (1/6) × (1/6) × (1/6) = 1/216.

Hence, the probability of getting 2 three times in a row is 0.463%.

问题 2. 连续两次掷出 1 的几率是多少？

解决方案：

Probability of an event = (number of favorable event) / (total number of event).

P(B) = (Event B) / (total number of event).

Probability of getting 1 = 1/6.

Rolling dice is an independent event, it is not dependent on how many times it’s been rolled.

Probability of getting 1 two times in a row = probability of getting 1 first time×probability of getting 1 second time.

Probability of getting 1 two times in a row  = (1/6) × (1/6) = 1/36.

Hence, the probability of getting 1 two times in a row 2.77 %.