问一道题：NO.PZ2017092702000082

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：

0.26

0.34

0.48

解释：

B is correct.

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

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

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

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

老师您好，看不太懂答案，可以解释一下具体解题步骤和计算器步骤吗？

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1 个答案

星星_品职助教 · 2020年05月17日

同学你好，

这道题问如果outperform的概率是60%，那么outperform的次数在五次试验中达到四次以上的概率是多少。

这是一道典型的二项分布的问题，由于要求5次里面outperform four or more times，就有两种情况，一种是outperform了4次，一种是5次。

以发生了4次为例，P(4)=5C4 * 60%^4 * (1-60%)^1，计算器可以直接算5C4： “5”， 2nd “+”，“4”， “=”。

同理算出P(5)，然后加总即可得到答案。

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欢欢 · 2021年11月09日

这道题为什么是组合而非排列呢？不用考虑顺序吗？排列和组合的概念还是没有非常清晰。

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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）

2023-08-16 02:49 1 · 回答

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.0.26B.0.34C.0.48B 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这道题计算没有错误么第一步得出来是0.5184呀C54是10*0.6^4*0.4=0.5184

2023-04-19 19:19 1 · 回答

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 求问公式里的( n x )在答案里是怎么体现出来的啊？

2022-07-19 07:25 1 · 回答

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！

2019-12-25 00:09 2 · 回答

问一道题：NO.PZ2017092702000082

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