Question #104193
What is the probability your friend will pass the exam. There are 20 true/false items on the exam. Each question has the same weight and 60% or more correct answers results in a passing grade.
1
Expert's answer
2020-03-02T11:28:28-0500

Consider XX  as the number of questions answered correctly among the 20 questions: XB(n,p)X\sim B(n,p)


P(X=x)=(nx)px(1p)nxP(X=x)=\binom{n}{x}p^x(1-p)^{n-x}

Given that p=0.5,n=20p=0.5, n=20

0.620=120.6\cdot20=12


P(X=12)=(2012)0.512(10.5)2012=0.120134P(X=12)=\binom{20}{12}0.5^12(1-0.5)^{20-12}=0.120134

P(X=13)=(2013)0.513(10.5)2013=0.073929P(X=13)=\binom{20}{13}0.5^{13}(1-0.5)^{20-13}=0.073929

P(X=14)=(2014)0.514(10.5)2014=0.036964P(X=14)=\binom{20}{14}0.5^{14}(1-0.5)^{20-14}=0.036964

P(X=15)=(2015)0.515(10.5)2015=0.014786P(X=15)=\binom{20}{15}0.5^{15}(1-0.5)^{20-15}=0.014786

P(X=16)=(2016)0.516(10.5)2016=0.004621P(X=16)=\binom{20}{16}0.5^{16}(1-0.5)^{20-16}=0.004621

P(X=17)=(2017)0.517(10.5)2017=0.001087P(X=17)=\binom{20}{17}0.5^{17}(1-0.5)^{20-17}=0.001087

P(X=18)=(2018)0.518(10.5)2018=0.000181P(X=18)=\binom{20}{18}0.5^{18}(1-0.5)^{20-18}=0.000181


P(X=19)=(2019)0.519(10.5)2019=0.000019P(X=19)=\binom{20}{19}0.5^{19}(1-0.5)^{20-19}=0.000019

P(X=20)=(2020)0.520(10.5)2020=0.000001P(X=20)=\binom{20}{20}0.5^ {20}(1-0.5)^{20-20}=0.000001

P(X12)=P(X=12)+P(X=13)+P(X=14)+P(X\geq12)=P(X=12)+P(X=13)+P(X=14)++P(X=15)+P(X=16)+P(X=17)+P(X=18)++P(X=15)+P(X=16)+P(X=17)+P(X=18)+

+P(X=19)+P(X=20)+P(X=19)+P(X=20)\approx 0.2517220.251722


The probability my friend will pass the exam is 0.25170.2517 (25%)(\approx25\%)



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