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ditto · 2023年11月05日

接下来3年大于1, 为什么只算p2和P3两种情况

NO.PZ2018062016000082

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:

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(2)=3!(32)!2!×0.32(10.3)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)=0.189.

p(3)=3!(33)!3!×0.33(10.3)0=(1)(0.027)(1)=0.027p(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\cdot

The required probability 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

2 个答案
已采纳答案

星星_品职助教 · 2023年11月06日

同学你好,

本题的概率是次数的概率。即题目要计算的“stock will rise more than 1 time”指的是超过1次的概率,故不需考虑仅一次的情况。

cfa考鸭 · 2023年11月05日

p1的所代表的是是

只有一次rise的情况的集合

题目问的是至少两次rise的情况

所以

答案不包含p1

ditto · 2023年11月06日

谢谢解答!

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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

2022-11-04 18:24 1 · 回答

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?

2022-08-21 16:46 1 · 回答

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?

2021-09-20 17:41 1 · 回答

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

2020-07-28 13:18 1 · 回答