开发者:上海品职教育科技有限公司 隐私政策详情

应用版本:4.2.11(IOS)|3.2.5(安卓)APP下载

saimeiei · 2019年07月11日

问一道题:NO.PZ2016062402000015

问题如下图:

    

选项:

A.

B.

C.

D.

解释:


老师好,请问一下这道题为什么不用泊松分布

1 个答案

品职答疑小助手雍 · 2019年07月11日

同学你好,当二项分布的n很大而p很小时,泊松分布可作为二项分布的近似。所以这题n不大,p也不小,最好还是不要近似。

把二项分布符合结果的情况列出来概率相加就行了。

  • 1

    回答
  • 0

    关注
  • 497

    浏览
相关问题

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

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%.请问为什么第二个计算要乘6但是第一个不用呀?

2020-10-22 18:04 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