Six questions test

There are six questions in the test. There are 3 answers to each - only one is correct. In order for a student to take the exam, at least four questions must be answered correctly. Alan didn't learn at all, 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:

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

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