掷骰子连续三次得到 6 的概率是多少？
概率是处理事件发生可能性的数学的一部分。就是预测事件发生或不发生的可能性有多大。作为数字的概率仅介于 0 和 1 之间，也可以写成百分比或分数的形式。可能事件 B 的概率通常写为 P(B)。这里 P 表示可能性，B 表示事件的发生。类似地，任何事件的概率通常写为 P()。当事件的最终结果未得到确认时，我们会使用某些结果的概率——它们发生的可能性或它们发生的机会。
虽然概率是从赌博开始的，但在物理科学、商业、生物科学、医学、天气预报等领域，它已经被谨慎地使用了。
为了更准确地理解概率，我们以掷骰子为例：
可能的结果是 - 1、2、3、4、5 和 6。
得到任何结果的概率是 1/6。由于事件发生的可能性是同等可能的事件，因此在这种情况下获得任何数字的可能性相同，它是 1/6 或 50/3%。
概率公式
Probability of an event = {Number of ways it can occur} ⁄ {Total number of outcomes}
P(A) = {Number of ways A occurs} ⁄ {Total number of outcomes}
活动类型
- 同等可能事件：掷骰子后，获得任何可能事件的概率为 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相同，它是公平掷骰子的 1/6。
- 补充事件：有可能只有两个结果，即一个事件是否会发生。就像一个人会玩或不玩，买笔记本电脑或不买笔记本电脑等都是互补事件的例子。
掷骰子连续三次得到 6 的概率是多少？
解决方案：-
Probability of an event = (number of favourable event) / (total number of event).
P(B) = (Event B) / (total number of event).
Probability of getting 6 = 1/6.
Rolling of an dice is an independent event, it is not dependent on how many times it’s been rolled.
Probability of getting 6 three times in a row = probability of getting 6 first time × probability of getting 6 second time × probability of getting 6 third time.
Probability of getting 6 three times in a row = (1/6) × (1/6) × (1/6) = 1/216.
Hence, the probability of getting 6 three times in a row is 0.463%.
类似问题
问题 1：骰子掷出 3 次 5 的概率是多少？
解决方案：
Probability of an event = (number of favourable event) / (total number of event).
P(B) = (Event B) / (total number of event).
Probability of getting 5 = 1/6.
Rolling of an dice is an independent event, it is not dependent on how many times it’s been rolled.
Probability of getting 5 three times in a row = probability of getting 5 first time × probability of getting 5 second time × probability of getting 5 third time.
Probability of getting 5 three times in a row = (1/6) × (1/6) × (1/6) = 1/216.
Hence, the probability of getting 5 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 favourable event) / (total number of event).
P(B) = (Event B) / (total number of event).
Probability of getting 1 = 1/6.
Rolling of an 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 %.