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

Correct answer:

p =  0.9999993

Step-by-step explanation:

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

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