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小壹万万呀 · 2020年10月22日

问一道题：NO.PZ2016062402000015 [ FRM I ]

  1. FRM I
  2. Quantitative

问题如下：

On a multiple-choice exam with four choices for each of six questions, what is the probability that a student gets fewer than two questions correct simply by guessing?

选项：

A.

0.46%

B.

23.73%

C.

35.60%

D.

53.39%

解释：

We use the density given by Equation: f(x)=(nx)px(1−p)n−xf{(x)}={(\begin{array}{l}n\\x\end{array})}p^x{(1-p)}^{n-x}. The number of trials is n = 6. The probability of guessing correctly just by chance is p = 1/4 = 0.25. The probability of zero lucky guesses is C600.2500.756=0.756=0.17798C_6^00.25^00.75^6=0.75^6=0.17798. The probability of one lucky guess is C610.2510.755=6×0.25×0.755=0.35596C_6^10.25^10.75^5=6\times0.25\times0.75^5=0.35596. The sum is 0.5339.

Note that the same analysis can be applied to the distribution of scores on an FRM examination with 100 questions. It would be virtually impossible to have a score of zero, assuming random guesses; this probability is 0.75100=3.2E−130.75^{100}=3.2E-13. Also, the expected percentage score under random guesses is p = 25%.

请问为什么第二个计算要乘6但是第一个不用呀？
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1 个答案

袁园_品职助教 · 2020年10月23日

因为 C6 0 =1，即六道题里面答对0道题只有一种可能

但是 C6 1 = 6，六道题里面答对任何一道题目是有6种可能的

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NO.PZ2016062402000015 问题如下 On a multiple-choiexwith four choices for eaof six questions, whis the probability tha stunt gets fewer thtwo questions corresimply guessing? 0.46% 23.73% 35.60% 53.39% We use the nsity given Equation: f(x)=(nx)px(1−p)n−xf{(x)}={(\begin{array}{l}n\\x\enarray})}p^x{(1-p)}^{n-x}f(x)=(nx​)px(1−p)n−x. The number of trials is n = 6. The probability of guessing correctly just chanis p = 1/4 = 0.25. The probability of zero lucky guesses is C600.2500.756=0.756=0.17798C_6^00.25^00.75^6=0.75^6=0.17798C60​0.2500.756=0.756=0.17798. The probability of one lucky guess is C610.2510.755=6×0.25×0.755=0.35596C_6^10.25^10.75^5=6\times0.25\times0.75^5=0.35596C61​0.2510.755=6×0.25×0.755=0.35596. The sum is 0.5339.Note ththe same analysis capplieto the stribution of scores on FRM examination with 100 questions. It woulvirtually impossible to have a score of zero, assuming ranm guesses; this probability is 0.75100=3.2E−130.75^{100}=3.2E-130.75100=3.2E−13. Also, the expectepercentage score unr ranm guesses is p = 25%. 老师好，请问这题中的二项式计算（下面）用计算器怎么按？谢谢！C60 *0.250*0.756

2022-05-03 16:33 1 · 回答

     老师好，请问一下这道题为什么不用泊松分布

2019-07-11 14:48 1 · 回答

     The sum is 0.5339，这一步是怎么得出的？

2018-09-08 13:04 1 · 回答

问一道题：NO.PZ2016062402000015 [ FRM I ] 这是一个多选题，每道题选对的概率应该是15分之1？

2017-05-07 20:55 1 · 回答