Question

Pat Statsdud is taking an economics course. Pat's exam strategy is to rely on luck for the next exam. The exam consists of 20 multiple-choice questions. Each question has four possible answers, only one of which is correct. Pat plans to guess the answer to each question without reading it. If a grade on the exam is 50% or more, Pat will pass the exam. Find the probability that Pat will pass the exam.

Answer 1

Answer:

0.01386 or 1.386%

Step-by-step explanation:

Each question has a binomial distribution with probability of success p =0.25 (1 correct answer out of four alternatives).

The probability of 'k' successes in n trials is given by:

$P(x=k)=\frac{n!}{(n-k)!k!}*p^k*(1-p)^{n-k}$

Pat will pass the exam if x ≥ 10. The probability that Pat will pass is:

$P(pass)=P(x=10)+P(x=11)+P(x=12)+P(x=13)+P(x=14)+P(x=15)+P(x=16)+P(x=17)+P(x=18)+P(x=19)+P(x=20)$

The probability for each number of success is:

$P(x=10)=\frac{20!}{(20-10)!10!}*0.25^{10}*0.75^{10}=0.0099\\\\P(x=11)= \frac{20!}{(20-11)!11!}*0.25^{11}*0.75^{9}=0.0030\\\\P(x=12)=\frac{20!}{(20-12)!12!}*0.25^{12}*0.75^{8}=0.00075\\\\P(x=13)=\frac{20!}{(20-13)!13!}*0.25^{13}*0.75^{7}=0.00015\\\\P(x=14)=\frac{20!}{(20-14)!14!}*0.25^{14}*0.75^{6}=0.0000257\\\\P(x=15)=\frac{20!}{(20-15)!15!}*0.25^{15}*0.75^{5}=3.426*10^{-6}$

$P(x=16)=\frac{20!}{(20-16)!16!}*0.25^{16}*0.75^{4}=3.569*10^{-7}\\\\P(x=17)=\frac{20!}{(20-17)!17!}*0.25^{17}*0.75^{3}=2.799*10^{-8}\\\\P(x=18)=\frac{20!}{(20-18)!18!}*0.25^{18}*0.75^{2}=1.555*10^{-9}\\\\P(x=19)=\frac{20!}{(20-19)!19!}*0.25^{19}*0.75^{1}=5.457*10^{-11}\\\\P(x=20)=\frac{20!}{(20-20)!20!}*0.25^{20}*0.75^{0}=9.095*10^{-13}$

The probability that Pat will pass his exam is:

$P(pass)=0.01386$