Internet anywhere

In school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that
a) exactly 5 have access to the internet.
b) At least 6 students have access to the internet

Correct result:

a =  0.2787
b =  0.3154

Solution:

$$\begin{gathered} C_5(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{5}}\right)={\displaystyle \frac{8!}{5!(8-5)!}}={\displaystyle \frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}}=56 \\ p=60\mathrm{/}100={\displaystyle \frac{3}{5}}=0.6 \\ q=1-p=1-0.6={\displaystyle \frac{2}{5}}=0.4 \\ a=\left(\genfrac{}{}{0px}{}{5}{8}\right)\cdot p^5\cdot q^{8-5}=\left(\genfrac{}{}{0px}{}{5}{8}\right)\cdot 0.6^5\cdot 0.4^{8-5}=0.2787 \end{gathered}$$

$$\begin{gathered} C_6(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{6}}\right)={\displaystyle \frac{8!}{6!(8-6)!}}={\displaystyle \frac{8\cdot 7}{2\cdot 1}}=28 \\ {C}_7(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{7}}\right)={\displaystyle \frac{8!}{7!(8-7)!}}={\displaystyle \frac{8}{1}}=8 \\ {C}_8(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{8}}\right)={\displaystyle \frac{8!}{8!(8-8)!}}={\displaystyle \frac{1}{1}}=1 \\ {p}_6=\left(\genfrac{}{}{0px}{}{6}{8}\right)\cdot p^6\cdot q^{8-6}=\left(\genfrac{}{}{0px}{}{6}{8}\right)\cdot 0.6^6\cdot 0.4^{8-6}\doteq 0.209 \\ {p}_7=\left(\genfrac{}{}{0px}{}{7}{8}\right)\cdot p^7\cdot q^{8-7}=\left(\genfrac{}{}{0px}{}{7}{8}\right)\cdot 0.6^7\cdot 0.4^{8-7}\doteq 0.0896 \\ {p}_8=\left(\genfrac{}{}{0px}{}{8}{8}\right)\cdot p^8\cdot q^{8-8}=1\cdot 0.6^8\cdot 0.4^{8-8}\doteq 0.0168 \\ b=p_6+p_7+p_8=0.209+0.0896+0.0168=0.3154 \end{gathered}$$

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