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Binomial Distribution Example

二项分布示例
Distribution
分布
Functional Form
函数形式
Mean
平均
Standard Deviation
标准差
Binomial
二项式

Joe throws a die six times and throws a three twice. It that incredibly lucky or unusual?

乔掷骰子六次,其中两次掷出三点。这是否异常幸运或罕见?

The basic probability of throwing a "3" in one throw is p=1/6 . Then for n=6 throws, the average number of 3's would be

单次掷骰子掷出'3'的概率是p=1/6。对于n=6次掷骰子,平均出现'3'的次数为

a=np= 6 x 1/6 = 1

a=np= 6 x 1/6 = 1(专名或术语)

So throwing two 3's is definitely not the expected or most probable result, but it remains to be seen how much the probability is diminished relative to the most probable result.

抛出两个3点 definitely 不是期望的结果,也不是最可能的结果,但还需看它相对于最可能结果的概率降低程度。

Now the number of ways to pick r objects out of a total of n is

现在,从n个物体中选出r个物体的方法数为

n(n-1)(n-2)...(n-r+1) = n!/(n-r)!

which is called the permutation. But considering them to be distinguishable, this overcounts. If getting the same objects or outcomes but in different order is not going to be considered to be a different overall outcome, then we need to divide by the number of ways to arrange them, which is r! . The number of ways to pick distinguishable sets is then

这被称为排列。但若将它们视为可区分的,这会重复计数。如果得到相同对象或结果但顺序不同不被视为不同的整体结果,那么就需要除以排列它们的方式数,即r!。因此,选取可区分集合的方式数为

n!/[r!(n-r)!]

which is called the combination.

这被称为组合。

The probability of getting a certain outcome r times is pr, but if you do that AND get some other outcome for the remainder of the tries (associated probability (1-p)n-r) then the probability for this combined outcome is the product of these probabilities. This is a part of the nature of general probability: the probability for any two outcomes related by a logical "AND" will be the product of the probabilities of the single outcomes. Then if the combined probability is multiplied by the number of ways to get this outcome, the result is the binomial distribution function.

得到特定结果r次的概率是p^r,但如果在这些尝试中同时得到其他结果(相关概率为(1-p)^{n-r}),则这种组合结果的概率是这两个概率的乘积。这是概率的一般性质的一部分:任何由逻辑“与”连接的两个结果的概率将是单个结果概率的乘积。然后,如果将这种组合概率乘以得到这种结果的方式数,结果就是二项分布函数。

Using this function to evaluate the probabilities of 0,1,2 and 3 times for throwing a "3" by throwing a die six times gives

利用这个函数计算掷骰子六次中出现0次、1次、2次和3次“3”的概率给
For 6 throws:
对于6次投掷:
Number of times:
次数:
Probability
概率
0
0.335
1
0.402
2
0.201
3
0.054

The probability for throwing a "3" twice is seen to be about half that for the expected or average value of one time. So it would not be unusual to see this happen, but it would be rarer than not throwing any 3's in six throws.

抛出两个‘3’的概率大约是单次抛掷期望值或平均值的一半。因此,看到这种情况发生并不罕见,但比六次抛掷中不出现‘3’的概率要小。
Calculation
计算
Comments about the form of the function
关于函数形式的评论
Example of throwing dice
掷骰子的例子
Index

Distribution functions

Applied statistics concepts
索引分布函数 应用统计学概念
 
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