红豆生南国 · 2019年11月09日
问题如下：
The stock of AAA company has a 30% probability to rise every year, if every annual trial is independent from each other, the probability that the stock will rise more than 1 time in the next 3 years is:
选项：
A.0.145
B.0.216
C.0.377
解释：
B is correct. Based on the corresponding formula:
, n = 3 and p = 0.30.
p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189\).
The required probability is: p(2) + p(3) = 0.189 + 0.027 = 0.216
这道题该用哪个公式解呢？
星星_品职助教 · 2019年11月09日
同学你好，
这道题是二项分布的考点。股票的变化只有两种可能，上涨或下跌，所以是个伯努利变量。做多次伯努利试验所产生的的成功“次数”的分布就是二项分布。所以如果看到求次数发生概率的题，就要联想到二项分布。
这道题问在三次试验中，股票上涨超过1次的概率。所以就是要么2次上涨，要么3次全上涨。就是P(x=2) + P(x=3)，直接代入二项分布的公式即可，这个公式是一定要会背的。以发生两次的概率为例，P(x=2)=3C2 * 0.3^2 * 0.7^1 = 0.1890. 其中3C2的部分不需要去记忆公式，直接按计算器。先输入3，然后 2nd “+”，然后输入2 求解即可。
加油。
NO.PZ2018062016000082 问题如下 The stoof Acompany ha 30% probability to rise every year, if every annutriis inpennt from eaother, the probability ththe stowill rise more th1 time in the next 3 years is: A.0.145 B.0.216 C.0.377 B 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(2)=3!(3−2)!2!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189p(2)=(3−2)!2!3!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189.p(3)=3!(3−3)!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅p(3)=\frac{3!}{(3-3)!3!}\times0.3^3(1-0.3)^0=\left(1\right)\left(0.027\right)\left(1\right)=0.027\ctp(3)=(3−3)!3!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅The requireprobability is: p(2) + p(3) = 0.189 + 0.027 = 0.216 我算的是接下来3年均大于， 也就是p1+p2+p3均大于的情况， 请问这样理解为什么不对
NO.PZ2018062016000082 问题如下 The stoof Acompany ha 30% probability to rise every year, if every annutriis inpennt from eaother, the probability ththe stowill rise more th1 time in the next 3 years is: A.0.145 B.0.216 C.0.377 B 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(2)=3!(3−2)!2!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189p(2)=(3−2)!2!3!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189.p(3)=3!(3−3)!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅p(3)=\frac{3!}{(3-3)!3!}\times0.3^3(1-0.3)^0=\left(1\right)\left(0.027\right)\left(1\right)=0.027\ctp(3)=(3−3)!3!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅The requireprobability is: p(2) + p(3) = 0.189 + 0.027 = 0.216 谢谢解答
NO.PZ2018062016000082问题如下The stoof Acompany ha 30% probability to rise every year, if every annutriis inpennt from eaother, the probability ththe stowill rise more th1 time in the next 3 years is:A.0.145B.0.216C.0.377B 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(2)=3!(3−2)!2!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189p(2)=(3−2)!2!3!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189.p(3)=3!(3−3)!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅p(3)=\frac{3!}{(3-3)!3!}\times0.3^3(1-0.3)^0=\left(1\right)\left(0.027\right)\left(1\right)=0.027\ctp(3)=(3−3)!3!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅The requireprobability is: p(2) + p(3) = 0.189 + 0.027 = 0.216老师，本题讲解没看懂，可否帮忙再详细一下呢？
NO.PZ2018062016000082 0.216 0.377 B 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(2)=3!(3−2)!2!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189p(2)=(3−2)!2!3!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189. p(3)=3!(3−3)!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅p(3)=\frac{3!}{(3-3)!3!}\times0.3^3(1-0.3)^0=\left(1\right)\left(0.027\right)\left(1\right)=0.027\ctp(3)=(3−3)!3!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅ The requireprobability is: p(2) + p(3) = 0.189 + 0.027 = 0.216求解未来三年至少还有一年上涨的概率是否可以想成排出三年来一次都不上涨的概率，即1-0.7*0.7*0.7。但是这种思路算出来的结果与标准答案大相径庭，错误之处在哪里呢？
0.216 0.377 B 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(2)=3!(3−2)!2!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189p(2)=\frac{3!}{(3-2)!2!}\times0.3^2(1-0.3)^1=\left(3\right)\left(0.09\right)\left(0.7\right)=0.189p(2)=(3−2)!2!3!×0.32(1−0.3)1=(3)(0.09)(0.7)=0.189. p(3)=3!(3−3)!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅p(3)=\frac{3!}{(3-3)!3!}\times0.3^3(1-0.3)^0=\left(1\right)\left(0.027\right)\left(1\right)=0.027\ctp(3)=(3−3)!3!3!×0.33(1−0.3)0=(1)(0.027)(1)=0.027⋅ The requireprobability is: p(2) + p(3) = 0.189 + 0.027 = 0.216老师您好 这道题的解题思路和所运用知识点您能帮忙解决一下吗
