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Nilinkat09 · 2020年03月16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.

老师,做题时没有想到二项式的考点,我理解成假设是正态分布,求F(X<=0.75)的累计概率,考虑到函数关于零点对称,那么0点左侧的部分概率是0.5,所以选择概率大于0.5的选项,请问这样的思路有问题吗?

1 个答案

星星_品职助教 · 2020年03月16日

同学你好,

这道题不能这样思考。因为这道题本质是二项分布,所以用正态分布来考虑肯定是有问题的。
二项分布的题干是很有特点的,当看到结果只有两种可能,且有成功概率(0.75),然后求发生次数的概率的时候,就是二项分布。

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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