问题如下图：
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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=0.17798C600.2500.756=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=0.35596C610.2510.755=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
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=0.17798C600.2500.756=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=0.35596C610.2510.755=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？
老师好，请问一下这道题为什么不用泊松分布
The sum is 0.5339，这一步是怎么得出的？