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010****0328 · 2021年02月20日

二项分布公示含义

NO.PZ2018062016000083

问题如下:

The stock of AAA company has a 30% probability to rise every year, assume that every annual trial is independent from each other. If the stock meets the goal of rising more than 1 time in the next 3 years, what is the probability that it fails to meet the goal?

选项:

A.

0.343

B.

0.216

C.

0.784

解释:

C is correct. Based on the corresponding formula:

p(x)=P(X=x)=(nx)px(1p)nxp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\end{array})}p^x{(1-p)}^{n-x}, n = 3 and p = 0.30.

p(1)=3!(31)!1!×0.31(10.3)2=(3)(0.3)(0.49)=0.441p(1)=\frac{3!}{(3-1)!1!}\times0.3^1(1-0.3)^2=\left(3\right)\left(0.3\right)\left(0.49\right)=0.441

p(0)=3!(30)!0!×0.30(10.3)3=(1)(1)(0.343)=0.343p(0)=\frac{3!}{(3-0)!0!}\times0.3^0(1-0.3)^3=\left(1\right)\left(1\right)\left(0.343\right)=0.343\cdot

The required probability is: p(1) + p(0) = 0.441 + 0.343 = 0.784

请问二项分布公式(n x)px (1-p)n-x 这个(n x)什么意思呢 
1 个答案

星星_品职助教 · 2021年02月20日

同学你好,

括号里的n和x代表是一共进行了n次试验,其中成功的有x次。这其实就是nCx的意思,nCx可以用计算器直接按出来。

以本题为例,n=3,如果要计算的是P(1),则此时x=1。3C1=3;

同理,P(0)中的x=0,此时3C0=1.

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NO.PZ2018062016000083 问题如下 The stoof Acompany ha 30% probability to rise every year, assume thevery annutriis inpennt from eaother. If the stomeets the goof rising more th1 time in the next 3 years, whis the probability thit fails to meet the goal? A.0.343 B.0.216 C.0.784 C is correct. Baseon the corresponng formula:p(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx​)px(1−p)n−x, n = 3 anp = 0.30.p(1)=3!(3−1)!1!×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441p(1)=\frac{3!}{(3-1)!1!}\times0.3^1(1-0.3)^2=\left(3\right)\left(0.3\right)\left(0.49\right)=0.441p(1)=(3−1)!1!3!​×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441p(0)=3!(3−0)!0!×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅p(0)=\frac{3!}{(3-0)!0!}\times0.3^0(1-0.3)^3=\left(1\right)\left(1\right)\left(0.343\right)=0.343\ctp(0)=(3−0)!0!3!​×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅The requireprobability is: p(1) + p(0) = 0.441 + 0.343 = 0.784 老师 能不能麻烦具体讲解一下这道题的思路 是要运用到nCr这个运算吗?

2023-03-08 09:52 1 · 回答

NO.PZ2018062016000083 0.216 0.784 C is correct. Baseon the corresponng formulp(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx​)px(1−p)n−x, n = 3 anp = 0.30. p(1)=3!(3−1)!1!×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441p(1)=\frac{3!}{(3-1)!1!}\times0.3^1(1-0.3)^2=\left(3\right)\left(0.3\right)\left(0.49\right)=0.441p(1)=(3−1)!1!3!​×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441 p(0)=3!(3−0)!0!×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅p(0)=\frac{3!}{(3-0)!0!}\times0.3^0(1-0.3)^3=\left(1\right)\left(1\right)\left(0.343\right)=0.343\ctp(0)=(3−0)!0!3!​×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅ The requireprobability is: p(1) + p(0) = 0.441 + 0.343 = 0.784 问的是失败的概率那不应该是成功两次和成功三次的反面吗。那不就是失败一次和失败0次,而失败是0.7啊怎么不用0.7算。我选的B a

2021-11-24 21:04 1 · 回答

NO.PZ2018062016000083 这道题表达很奇怪啊 看不懂 说如果meet了上涨超过一次的goal,那么不meet goal的几率是多少 这表达完全错误啊 根本看不懂

2021-07-02 23:34 1 · 回答

NO.PZ2018062016000083 0.216 0.784 C is correct. Baseon the corresponng formulp(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\enarray})}p^x{(1-p)}^{n-x}p(x)=P(X=x)=(nx​)px(1−p)n−x, n = 3 anp = 0.30. p(1)=3!(3−1)!1!×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441p(1)=\frac{3!}{(3-1)!1!}\times0.3^1(1-0.3)^2=\left(3\right)\left(0.3\right)\left(0.49\right)=0.441p(1)=(3−1)!1!3!​×0.31(1−0.3)2=(3)(0.3)(0.49)=0.441 p(0)=3!(3−0)!0!×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅p(0)=\frac{3!}{(3-0)!0!}\times0.3^0(1-0.3)^3=\left(1\right)\left(1\right)\left(0.343\right)=0.343\ctp(0)=(3−0)!0!3!​×0.30(1−0.3)3=(1)(1)(0.343)=0.343⋅ The requireprobability is: p(1) + p(0) = 0.441 + 0.343 = 0.784 目标是大于等于1次,fail完成目标就应该是小于0次,就应该只计算p(0)的概率啊

2021-05-07 13:01 1 · 回答