Records

Records indicate 90% error-free. If 8 records are randomly selected, what is the probability that at least 2 records have no errors?

Correct answer:

p =  0.9999993

Step-by-step explanation:

$$\begin{gathered} q=90\mathrm{\%}={\displaystyle \frac{90}{100}}=0.9 \\ {C}_2(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{2}}\right)={\displaystyle \frac{8!}{2!(8-2)!}}={\displaystyle \frac{8\cdot 7}{2\cdot 1}}=28 \\ {p}_2=\left(\genfrac{}{}{0px}{}{8}{2}\right)\cdot q^2\cdot (1-q{)}^{8-2}=28\cdot 0.9^2\cdot (1-0.9{)}^{8-2}\doteq 2.268\cdot 10^{-5} \\ {C}_3(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{3}}\right)={\displaystyle \frac{8!}{3!(8-3)!}}={\displaystyle \frac{8\cdot 7\cdot 6}{3\cdot 2\cdot 1}}=56 \\ {p}_3=\left(\genfrac{}{}{0px}{}{8}{3}\right)\cdot q^3\cdot (1-q{)}^{8-3}=56\cdot 0.9^3\cdot (1-0.9{)}^{8-3}\doteq 0.0004 \\ {C}_4(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{4}}\right)={\displaystyle \frac{8!}{4!(8-4)!}}={\displaystyle \frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}}=70 \\ {p}_4=\left(\genfrac{}{}{0px}{}{8}{4}\right)\cdot q^4\cdot (1-q{)}^{8-4}=70\cdot 0.9^4\cdot (1-0.9{)}^{8-4}\doteq 0.0046 \\ {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}_5=\left(\genfrac{}{}{0px}{}{8}{5}\right)\cdot q^5\cdot (1-q{)}^{8-5}=56\cdot 0.9^5\cdot (1-0.9{)}^{8-5}\doteq 0.0331 \\ {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 \\ {p}_6=\left(\genfrac{}{}{0px}{}{8}{6}\right)\cdot q^6\cdot (1-q{)}^{8-6}=28\cdot 0.9^6\cdot (1-0.9{)}^{8-6}\doteq 0.1488 \\ {C}_7(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{7}}\right)={\displaystyle \frac{8!}{7!(8-7)!}}={\displaystyle \frac{8}{1}}=8 \\ {p}_7=\left(\genfrac{}{}{0px}{}{8}{7}\right)\cdot q^7\cdot (1-q{)}^{8-7}=8\cdot 0.9^7\cdot (1-0.9{)}^{8-7}\doteq 0.3826 \\ {C}_8(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{8}}\right)={\displaystyle \frac{8!}{8!(8-8)!}}={\displaystyle \frac{1}{1}}=1 \\ {p}_8=\left(\genfrac{}{}{0px}{}{8}{8}\right)\cdot q^8\cdot (1-q{)}^{8-8}=1\cdot 0.9^8\cdot (1-0.9{)}^{8-8}\doteq 0.4305 \\ p=p_2+p_3+p_4+p_5+p_6+p_7+p_8=2.268\cdot 10^{-5}+0.0004+0.0046+0.0331+0.1488+0.3826+0.4305=0.9999993 \end{gathered}$$

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