问题如下图：
选项：
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解释：
老师你好。请问一下p(2)=C32X0.3²X0.7）为什么0.7前不乘C31
菲菲_品职助教 · 2018年07月18日
同学你好~
首先你要理解一下二项分布的这个公式的原理，这部分你可以再回去听一下Binomial Distribution的老师讲的内容~
在这道题目中，比如要求p(2),相当于求三年中两年股价都上涨的概率，根据二项分布的公式，就等于C32(三年中任意抽取两年股价上涨，不排序)乘以0.3²(三年内两年股价上涨的概率)再乘以0.7(三年内一年股价不上涨的概率）。
加油~~
六粒橙 · 2018年07月19日
嗯我想起来了 谢谢老师！
菲菲_品职助教 · 2018年07月19日
嘻嘻，不用客气的，是我们应该做的~加油加油~有问题及时问哦~~
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