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Hugo(Xie Lanzhi) · 2019年12月25日

问一道题：NO.PZ2017092702000082

  1. CFA I
  2. Quantitative

问题如下：

A portfolio manager annually outperforms her benchmark 60% of the time. Assuming independent annual trials, what is the probability that she will outperform her benchmark four or more times over the next five years?

选项：

A.

0.26

B.

0.34

C.

0.48

解释：

B is correct.

To calculate the probability of 4 years of outperformance, use the formula: p(x)=P(X=x)=(nx)px(1−p)n−xp{(x)}=P{(X=x)}={(\begin{array}{c}n\\x\end{array})}p^x{(1-p)}^{n-x}  Using this formula to calculate the probability in 4 of 5 years, n = 5, x = 4 and p = 0.60. Therefore,

p(4)=5!(5−4)!4!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p{(4)}=\frac{5!}{{(5-4)}!4!}0.6^4{(1-0.6)}^{5-4}={(120/24)}{(0.1296)}{(0.40)}=0.2592

p(5)=5!(5−4)!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778p{(5)}=\frac{5!}{{(5-4)}!5!}0.6^5{(1-0.6)}^{5-5}={(120/120)}{(0.0778)}{(1)}=0.0778

The probability of outperforming 4 or more times is p(4) + p(5) = 0.2592 + 0.0778 = 0.3370

不懂公式，尤其是感叹号那块儿，求讲解，以及如何一步一步的按计算器，以及困扰了两天了。谢谢！

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星星_品职助教 · 2019年12月25日

同学你好，

这道题和你之前问的一样，都是二项分布的问题。这道题稍微复杂一些，因为问的是5次里面的four or more times，也就是说有两种情况，一种是4次，一种是5次，都算出来加和即可。
以发生了4次为例，P(4)=5C4 * 60%^4 * (1-60%)^1，答案解析公式里那些阶乘“！”是对5C4的展开，可以忽略，因为计算器可以直接算5C4.

同上一道题一样，5C4按顺序按计算器即可  “5”， 2nd “+”，“4”， “=”。

以上过程可以复习一下视频里的二项分布知识点，如果还有不明白的可以追问~加油。



Viola欧拉 · 2022年06月16日

老师 C这个公式貌似中学数学，忘得一干二净了，请问这个概念叫啥名字，我去搜下B站重学一下，加强理解

星星_品职助教 · 2022年06月16日

@杨茗 Viola

“C”是“Combination”也就是组合，这不是重点考点，会用计算器按出结果就可以了。中学学的排列组合难度比这里大很多。

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NO.PZ2017092702000082 问题如下 A portfolio manager annually outperforms her benchmark 60% of the time. Assuming inpennt annutrials, whis the probability thshe will outperform her benchmark four or more times over the next five years? A.0.26 B.0.34 C.0.48 B is correct.To calculate the probability of 4 years of outperformance, use the 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 Using this formula to calculate the probability in 4 of 5 years, n = 5, x = 4 anp = 0.60. Therefore, p(4)=5!(5−4)!4!0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p{(4)}=\frac{5!}{{(5-4)}!4!}0.6^4{(1-0.6)}^{5-4}={(120/24)}{(0.1296)}{(0.40)}=0.2592p(4)=(5−4)!4!5!​0.64(1−0.6)5−4=(120/24)(0.1296)(0.40)=0.2592p(5)=5!(5−4)!5!0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778p{(5)}=\frac{5!}{{(5-4)}!5!}0.6^5{(1-0.6)}^{5-5}={(120/120)}{(0.0778)}{(1)}=0.0778p(5)=(5−4)!5!5!​0.65(1−0.6)5−5=(120/120)(0.0778)(1)=0.0778The probability of outperforming 4 or more times is p(4) + p(5) = 0.2592 + 0.0778 = 0.3370 计算outperform四次的公式如下5C4*0.6^4为什么不是后面乘以5C1（1-0.6），而是只乘以1（1-0.6）

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2023-04-19 19:19 1 · 回答

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老师您好，看不太懂答案，可以一下具体解题步骤和计算器步骤吗？

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