应该只计算p（0）吧？

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：

0.343

0.216

0.784

解释：

C is correct. Based on the corresponding formula:

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

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

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

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

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

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1 个答案

星星_品职助教 · 2021年05月07日

同学你好，

注意题干中的描述“...he goal of rising more than 1 time.....”，这个描述（more than）的意思是大于1，而不是大于等于1.

所以goal failed的情况就包含了1）rise 1 time；2）rise 0 time。

将这两种情况分别算出概率后相加即可。

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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请问二项分布公式（n x）px （1-p）n-x 这个（n x）什么意思呢

2021-02-20 14:15 1 · 回答

应该只计算p（0）吧？

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