Pass a test

The student has to pass a test that contains ten questions. For each of them, he chooses one of 5 answers, with just one being correct. The student did not prepare for the test, so he randomly chose the answers. What are the probabilities that the student will answer correctly:
a) a maximum of 3 questions
b) at least 4 questions
c) all ten questions

Correct answer:

a =  0.8791
b =  0.1209
c =  0

Step-by-step explanation:

$$\begin{gathered} n=10 \\ p=1/5=\frac{1}{5}=0.2 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{10}{0})\cdot 0.2^0\cdot (1-0.2{)}^{10-0}=1\cdot 0.2^0\cdot 0.8^{10}\doteq 0.1074 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{10}{1})\cdot 0.2^1\cdot (1-0.2{)}^{10-1}=10\cdot 0.2^1\cdot 0.8^9\doteq 0.2684 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{10}{2})\cdot 0.2^2\cdot (1-0.2{)}^{10-2}=45\cdot 0.2^2\cdot 0.8^8\doteq 0.302 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{10}{3})\cdot 0.2^3\cdot (1-0.2{)}^{10-3}=120\cdot 0.2^3\cdot 0.8^7\doteq 0.2013 \\ a=p_0+p_1+p_2+p_3=0.1074+0.2684+0.302+0.2013=0.8791 \end{gathered}$$

$b=1-a=1-0.8791=0.1209$

$c=\left(\genfrac{}{}{0ex}{}{n}{10}\right)\cdot p^{10}\cdot (1-p{)}^{n-10}=(\genfrac{}{}{0ex}{}{10}{10})\cdot 0.2^{10}\cdot (1-0.2{)}^{10-10}=1\cdot 0.2^{10}\cdot 0.8^0=0$

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