如果一个骰子掷了 5 次，掷出小于 3 的数字至少 3 次的概率是多少？

简单来说，概率就是研究一个特定事件相对于许多其他可能事件发生的可能性，其中不可能有两个或多个同时发生，即在给定的时间点只有一个可能发生的事件。概率应用的最简单示例是确定出现 1/2 头的可能性，即。有利的结果（头部出现）除以可能的结果（头部和尾部）。

二项分布

让我们继续掷硬币的相同活动，现在假设你的朋友建议你掷硬币 3 次，如果至少出现一次头像，你必须向他扔零食。但是，你知道你只剩下很少的钱可以花。在这里，您有必要了解您被迫扔零食的可能性是多少。

在这种情况下，在独立试验中涉及成功和失败的情况下，用于计算概率的过程称为二项分布法或伯努利分布法。

它之所以有这样的名字，是因为它只涉及两个可能的事件，即成功和失败，它们分布在执行的试验中。

为了使用此方法，应牢记以下几点：
应该有固定和有限的试验次数。
在每次试验中，成功和失败的概率应该是恒定的。
每个试验都应独立于其他试验。

如果同时满足这些点，则允许使用二项分布方法。现在回到您朋友设置的条件，我们可以看到应用二项式分布方法所需的所有条件都已满足。这是一个实际问题，现在让我们解决一个二项分布的理论问题，以便更好地理解它。

让我们假设有一个无偏的六面骰子，并在总共 5 次骰子中找到至少 3 次小于 3 的数字的概率。

基本排列原则

为了更好地理解有关二项分布的问题，重要的是要理解将事物连续排列的概念。可能有许多方法可以安排相同或不同的事物，有时计算这些安排变得很重要。如果有 (n + m) 个事物，其中 n 与一种相同，而 m 与另一种相同，则将它们排成一行的总方式由 (n + m)!/(m! × n！）。

如果一个骰子掷了 5 次，掷出小于 3 的数字至少 3 次的概率是多少？

解决方案：

Steps to proceed

According to the question, our favorable outcome is ⇢ Among the 5 independent rolls, there maybe 3 or 4 or all 5 numbers appearing less than 3.
Consider the favorable outcome as the success event(S) and calculate its probability, i.e. P(S) = 2/6. Hence out failure event(F) has a probability of, P(F)= 1 – P(S) = 4/6.
Now just to draw a similarity with the Basic principle of arrangement concept, that we will be using soon, let us consider the probability of success as a thing of one kind and the probability of failure P(F) as a thing of another kind non-identical from P(S). There may be three possible cases:

Event E₁– 3 numbers less than 3 and 2 numbers greater than or equal to 3. P(E₁) = (5!/(3! × 2!)) × (P(S))³ × (P(F))²

= (5!/(3! × 2!)) × (1/3)³ × (2/3)²

= 40/243

Event E₂ – 4 numbers less than 3 and 1 number greater than or equal to 3. P(E₂) = (5!/(4! × 1!)) × (P(S))⁴ × (P(F))

= (5!/(4! × 1!)) × (1/3)⁴ × (2/3)

= 10/243

Event E₃ – 5 numbers less than 3. p(E₃) = (5!/(5! × 0!)) × (P(S))⁵ × (P(F))⁰

= (1/3))⁵ × (2/3)⁰

= 1/243

Use the formula of the basic principle of combination because, here all successful events can be considered as identical and the failure events as identical but of a different kind, hence the number of ways of their arrangement in a row can be counted by using the basic principle of combination. For the final answer we need the union of all the three possibilities, i.e., (a) or (b) or (c). Let the final probability be P(Q) = (a) + (b) + (c)

= 40/243 + 10/243 + 1/243

= 51/243

编程需要懂一点英语

示例示例

问题1：抛硬币3次，尾巴只出现一次的概率是多少？

解决方案：

Let the probability of getting a head be P(H) = 1/2 and that of getting a tail to be P(T) = 1/2

Let the event of our favorable outcome, that is tail will appear only once out of the three performed trials be P(E)

All the conditions to apply the Binomial Distribution formula are fulfilled, therefore,

P(E) = 3!/(1! × 2!) × (P(H))² × (P(T)); if the tail has to appear only once the head has to appear twice.

= 3!/(1! × 2!) × (1/2))² × (1/2)

= 3/8

编程需要懂一点英语

问题 2：一个人能以 20% 的概率击中靶心。如果进行 4 次试验，他最多可以击中 3 次的概率是多少？

解决方案：

Let the probability of hitting the target be P(H) = 20/100 = 1/5 and that of failing to hit it be P(FH) = 1 – P(H)

= 4/5

Our favorable outcome is to get 3, 2 ,1 or no hits at all in the four trials, let it be P(F)

All conditions to apply Binomial Distribution formula are fulfilled, therefore;

If he fails to hit every time then; P1 = 4!/4! × (P(FH))⁴

= (4/5)⁴

= 256/625

If he hits it once; P₂ = 4!/(1! × 3!) × (P(H)) × (P(FH))³

= 4!/(1! × 3!) × (1/5) × (4/5))³

= 256/625

If he hits it twice; P₃ = 4!/(2! × 2!) × (P(H))² × (P(FH))²

= 4!/(2! × 2!) × (1/5)² × (4/5)²

= 96/625

If he hits it thrice; P₄= 4!/(3! × 1!) × (P(H))³ × (P(FH))

= 4!/(3! × 1!) × (1/5)³ × (4/5)

= 16/625

Therefore, P(F) = P₁ + P₂ + P₃ + P₄
= 624/625

编程需要懂一点英语

问题 3：Rakesh 和 Simi 两个朋友一个接一个地掷骰子。他们决定只在其中一个得到偶数时停止。可以通过二项分布法得到西米获胜的概率吗？为什么/为什么不？

解决方案：

Let the probability of getting an odd number on an unbiased dice be P(O) = 1/2 and that of getting an even number be P(E) = 1/2

Verifying required conditions:

Each trial by either friend is independent of all other trials
Only two events are possible one is the success and the other failure
With each trial the probability of success and failure does not change
There may be an infinite number of trials

The 4th point violates the conditions required to apply the Binomial Distribution method, because it may be possible that an even number appeared at the nth trial where n tends to infinity.

Hence the question cannot be solved by the Binomial Distribution method.

编程需要懂一点英语