Six questions test

There are six questions in the test. There are three answers to each - only one is correct. To take the exam, students must answer at least four questions correctly. Alan didn't learn, so he circled the answers only by guessing. What is the probability that Alan will not pass the exam?

Correct answer:

p =  0.8999

Step-by-step explanation:

$$\begin{gathered} q=\frac{1}{3}\doteq 0.3333 \\ {C}_4(6)=\left(\genfrac{}{}{0ex}{}{6}{4}\right)=\frac{6!}{4!(6-4)!}=\frac{6\cdot 5}{2\cdot 1}=15 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot q^4\cdot (1-q{)}^2=15\cdot {\frac{1}{3}}^4\cdot (1-\frac{1}{3}{)}^2=\frac{20}{243}\doteq 0.0823 \\ {C}_5(6)=\left(\genfrac{}{}{0ex}{}{6}{5}\right)=\frac{6!}{5!(6-5)!}=\frac{6}{1}=6 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{6}{5}\right)\cdot q^5\cdot (1-q{)}^1=6\cdot {\frac{1}{3}}^5\cdot (1-\frac{1}{3}{)}^1=\frac{4}{243}\doteq 0.0165 \\ {C}_6(6)=\left(\genfrac{}{}{0ex}{}{6}{6}\right)=\frac{6!}{6!(6-6)!}=\frac{1}{1}=1 \\ {p}_6=\left(\genfrac{}{}{0ex}{}{6}{6}\right)\cdot q^6\cdot (1-q{)}^0=1\cdot {\frac{1}{3}}^6\cdot (1-\frac{1}{3}{)}^0=\frac{1}{729}\doteq 0.0014 \\ p=1-p_4-p_5-p_6=1-\frac{20}{243}-\frac{4}{243}-p_6=1-0.0823-0.0165-0.0014=\frac{656}{729}=0.8999 \end{gathered}$$

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