aiyy0203. · 2023年03月08日
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:
, n = 3 and p = 0.30.
The required probability 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？
星星_品职助教 · 2023年03月08日
同学你好，
这道题是二项分布的一个考点。题干中问没有rise more than 1 time的概率，就相当于仅“rise”了1次或者0次的概率。这就是p(1) + p(0)
此后就是分别算出p(1) 和p(0)，分别代入二项分布的公式即可。
（上述公式中，括号内即是 nCx 的同义表述）
以仅rise一次为例。相当于同时满足两个条件：1. 三年中有一年rise； 2. 三年中的另外两年都是not rise。所以概率不仅要包括成功那一次的0.3，还要包括另外两年都要失败的概率0.7的平方，才能满足“仅”rise一次的概率。所以是3C1×(0.3^1)×(0.7^2)=0.441.
同理，rise 0次的概率是3C0 × (0.3^0) × (0.7)^3=0.343.
最终，没有完成目标的总概率就是以上两者相加。
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
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
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
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
