问一道题：NO.PZ2017092702000083 [ CFA I ]

问题如下：

If the probability that a portfolio outperforms its benchmark in any quarter is 0.75, the probability that the portfolio outperforms its benchmark in three or fewer quarters over the course of a year is closest to:

选项：

0.26

0.42

0.68

解释：

C is correct.

The probability that the performance is at or below the expectation is calculated by finding F(3) = p(3) + p(2) + p(1) + p(0) using the formula:

$p{(3)}=\frac{4!}{{(4-3)}!3!}0.75^3{(1-0.75)}^{4-3}={(24/6)}{(0.42)}{(0.25)}=0.42$

$p{(2)}=\frac{4!}{{(4-2)}!2!}0.75^2{(1-0.75)}^{4-2}={(24/4)}{(0.56)}{(0.06)}=0.20$

$p{(1)}=\frac{4!}{{(4-3)}!1!}0.75^1{(1-0.75)}^{4-1}={(24/6)}{(0.75)}{(0.02)}=0.06$

$p{(0)}=\frac{4!}{{(4-0)}!0!}0.75^0{(1-0.75)}^{4-0}={(24/24)}{(1)}{(0.004)}=0.004$

Therefore

F(3) = p(3) + p(2) + p(1) + p(0)= 0.42 + 0.20 + 0.06 + 0.004= 0.684 or approximately 68 percent.

这道题不太懂可以详细讲一下吗？是什么知识点

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1 个答案

星星_品职助教 · 2020年03月09日

同学你好，

这道题考察二项分布的公式。可以理解为成功()outperforms its benchmark)的概率是0.75，失败概率是0.25，然后求4次里成功3次或更少次数的概率。直接代入二项分布的公式计算即可。

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NO.PZ2017092702000083 问题如下 If the probability tha portfolio outperforms its benchmark in any quarter is 0.75, the probability ththe portfolio outperforms its benchmark in three or fewer quarters over the course of a yeis closest to: A.0.26 B.0.42 C.0.68 C is correct.The probability ththe performanis or below the expectation is calculatefinng F(3) = p(3) + p(2) + p(1) + p(0) using the formula:p(3)=4!(4−3)!3!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p{(3)}=\frac{4!}{{(4-3)}!3!}0.75^3{(1-0.75)}^{4-3}={(24/6)}{(0.42)}{(0.25)}=0.42p(3)=(4−3)!3!4!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p(2)=4!(4−2)!2!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p{(2)}=\frac{4!}{{(4-2)}!2!}0.75^2{(1-0.75)}^{4-2}={(24/4)}{(0.56)}{(0.06)}=0.20p(2)=(4−2)!2!4!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p(1)=4!(4−3)!1!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p{(1)}=\frac{4!}{{(4-3)}!1!}0.75^1{(1-0.75)}^{4-1}={(24/6)}{(0.75)}{(0.02)}=0.06p(1)=(4−3)!1!4!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p(0)=4!(4−0)!0!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004p{(0)}=\frac{4!}{{(4-0)}!0!}0.75^0{(1-0.75)}^{4-0}={(24/24)}{(1)}{(0.004)}=0.004p(0)=(4−0)!0!4!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004ThereforeF(3) = p(3) + p(2) + p(1) + p(0)= 0.42 + 0.20 + 0.06 + 0.004= 0.684 or approximately 68 percent. 这道题要求的是outperforms its benchmark in“ three or fewer quarters ”。所以包括了两种情况①正好有3个季度超过benchmark②超过benchmark的季度数少于3个季度（2个，1个和0个季度）所以列式是 p(3) + [p(2) + p(1) + p(0)]计算只涉及到二项分布的公式。本题中，一共四个季度，所以n=4；成功概率为0.75，所以失败概率就是1-0.75=0.25以P(3)也就是成功3次（x=3）为例此时相当于4次里面成功（outperform）了3次，还有1次就必须不成功。代入公式4C3× 0.75^3 × 0.25^1=0.4219 能或者翻译一下吗，看不出来是二项分布

2022-11-26 20:17 1 · 回答

NO.PZ2017092702000083问题如下 If the probability tha portfolio outperforms its benchmark in any quarter is 0.75, the probability ththe portfolio outperforms its benchmark in three or fewer quarters over the course of a yeis closest to:A.0.26 B.0.42C.0.68 C is correct.The probability ththe performanis or below the expectation is calculatefinng F(3) = p(3) + p(2) + p(1) + p(0) using the formula:p(3)=4!(4−3)!3!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p{(3)}=\frac{4!}{{(4-3)}!3!}0.75^3{(1-0.75)}^{4-3}={(24/6)}{(0.42)}{(0.25)}=0.42p(3)=(4−3)!3!4!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p(2)=4!(4−2)!2!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p{(2)}=\frac{4!}{{(4-2)}!2!}0.75^2{(1-0.75)}^{4-2}={(24/4)}{(0.56)}{(0.06)}=0.20p(2)=(4−2)!2!4!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p(1)=4!(4−3)!1!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p{(1)}=\frac{4!}{{(4-3)}!1!}0.75^1{(1-0.75)}^{4-1}={(24/6)}{(0.75)}{(0.02)}=0.06p(1)=(4−3)!1!4!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p(0)=4!(4−0)!0!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004p{(0)}=\frac{4!}{{(4-0)}!0!}0.75^0{(1-0.75)}^{4-0}={(24/24)}{(1)}{(0.004)}=0.004p(0)=(4−0)!0!4!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004ThereforeF(3) = p(3) + p(2) + p(1) + p(0)= 0.42 + 0.20 + 0.06 + 0.004= 0.684 or approximately 68 percent. 这道题要求的是outperforms its benchmark in“ three or fewer quarters ”。所以包括了两种情况①正好有3个季度超过benchmark②超过benchmark的季度数少于3个季度（2个，1个和0个季度）所以列式是 p(3) + [p(2) + p(1) + p(0)]计算只涉及到二项分布的公式。本题中，一共四个季度，所以n=4；成功概率为0.75，所以失败概率就是1-0.75=0.25以P(3)也就是成功3次（x=3）为例此时相当于4次里面成功（outperform）了3次，还有1次就必须不成功。代入公式4C3× 0.75^3 × 0.25^1=0.4219 老师，我根据公式，得出来了0.42，为什么是选择0.68呢？

2022-07-06 13:15 1 · 回答

NO.PZ2017092702000083 问题如下 If the probability tha portfolio outperforms its benchmark in any quarter is 0.75, the probability ththe portfolio outperforms its benchmark in three or fewer quarters over the course of a yeis closest to: A.0.26 B.0.42 C.0.68 C is correct.The probability ththe performanis or below the expectation is calculatefinng F(3) = p(3) + p(2) + p(1) + p(0) using the formula:p(3)=4!(4−3)!3!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p{(3)}=\frac{4!}{{(4-3)}!3!}0.75^3{(1-0.75)}^{4-3}={(24/6)}{(0.42)}{(0.25)}=0.42p(3)=(4−3)!3!4!0.753(1−0.75)4−3=(24/6)(0.42)(0.25)=0.42p(2)=4!(4−2)!2!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p{(2)}=\frac{4!}{{(4-2)}!2!}0.75^2{(1-0.75)}^{4-2}={(24/4)}{(0.56)}{(0.06)}=0.20p(2)=(4−2)!2!4!0.752(1−0.75)4−2=(24/4)(0.56)(0.06)=0.20p(1)=4!(4−3)!1!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p{(1)}=\frac{4!}{{(4-3)}!1!}0.75^1{(1-0.75)}^{4-1}={(24/6)}{(0.75)}{(0.02)}=0.06p(1)=(4−3)!1!4!0.751(1−0.75)4−1=(24/6)(0.75)(0.02)=0.06p(0)=4!(4−0)!0!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004p{(0)}=\frac{4!}{{(4-0)}!0!}0.75^0{(1-0.75)}^{4-0}={(24/24)}{(1)}{(0.004)}=0.004p(0)=(4−0)!0!4!0.750(1−0.75)4−0=(24/24)(1)(0.004)=0.004ThereforeF(3) = p(3) + p(2) + p(1) + p(0)= 0.42 + 0.20 + 0.06 + 0.004= 0.684 or approximately 68 percent. 这道题要求的是outperforms its benchmark in“ three or fewer quarters ”。所以包括了两种情况①正好有3个季度超过benchmark②超过benchmark的季度数少于3个季度（2个，1个和0个季度）所以列式是 p(3) + [p(2) + p(1) + p(0)]计算只涉及到二项分布的公式。本题中，一共四个季度，所以n=4；成功概率为0.75，所以失败概率就是1-0.75=0.25以P(3)也就是成功3次（x=3）为例此时相当于4次里面成功（outperform）了3次，还有1次就必须不成功。代入公式4C3× 0.75^3 × 0.25^1=0.4219 请问能不能这样算，任一季度超过benchmark的概率是0.75，要求至多三个季度超过benchmark的概率，等于1减去四个季度都超过benchmark的概率，就是1-0.75*4=0.6835

2022-05-17 14:02 1 · 回答

NO.PZ2017092702000083 这道题怎么算的？？我完全看不懂。。。。太复杂了

2021-02-05 15:38 2 · 回答

请问老师，为啥4c1*0.75^1*0.25^3不对呢。。我可能没读懂题。麻烦您详细讲解一下可以么

2020-12-13 23:13 2 · 回答

问一道题：NO.PZ2017092702000083 [ CFA I ]

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