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Hugo(Xie Lanzhi) · 2020年04月10日

问一道题:NO.PZ2017092702000083

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:

A.

0.26

B.

0.42

C.

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)=4!(43)!3!0.753(10.75)43=(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)}=0.42

p(2)=4!(42)!2!0.752(10.75)42=(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)}=0.20

p(1)=4!(43)!1!0.751(10.75)41=(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)}=0.06

p(0)=4!(40)!0!0.750(10.75)40=(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)}=0.004

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.

请问计算器要如何操作呢?

5 个答案

星星_品职助教 · 2020年04月27日

P(4)指的是四个季度全部outperform的概率,二项分布这个知识点重新学一下吧

星星_品职助教 · 2020年04月14日

你算的对,就是这么算的,要有信心。

可以把F(3) = p(3) + p(2) + p(1) + p(0)自己算一下,就是0.6836.

上课讲过的方法是F(3)=1-P(4),这么算更简单,结果不变。

星星_品职助教 · 2020年04月13日

二项公式前面的那部分直接用计算器计算combination的功能就可以,不需要写成阶乘的形式。如果是 4C2×0.75^2×0.25^2,结果就是0.2109。

答案解析只需要看一下过程就可以发现,是逐项计算且逐项四舍五入保留两位小数的,而不是一个公式直接算出来的。所以有小幅差异很正常。

Hugo(Xie Lanzhi) · 2020年04月14日

还是不太懂,我算的:4C1×0.75^1×0.25^3=0.0469,和答案相差挺多的。这是怎么回事呢?

星星_品职助教 · 2020年04月12日

例如4C2,按法就是“4”,2nd “+”, “2”,然后按“=”就可以得到6的结果。

这些上课都讲过,复习即可

星星_品职助教 · 2020年04月10日

同学你好,

没有计算器的快捷算法,按照二项分布公式正常计算即可。

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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