Electricity

Many companies providing public utility services promote energy saving by offering discount rates for consumers who keep their energy consumption below certain set standards. A recent EPA report states that 70% of the inhabitants of a monitored region reduced their consumption of electrical energy sufficiently to qualify for discounted rates. If ten consumers are randomly selected from this region, what is the probability that at least four persons will qualify for preferential prices?

Final Answer:

p =  0.9894

Step-by-step explanation:

$$\begin{gathered} k=0.70 \\ q=1-k=1-0.7=\frac{3}{10}=0.3 \\ {C}_0(10)=\left(\genfrac{}{}{0ex}{}{10}{0}\right)=\frac{10!}{0!(10-0)!}=\frac{1}{1}=1 \\ {b}_0=\left(\genfrac{}{}{0ex}{}{10}{0}\right)\cdot k^0\cdot q^{10}=1\cdot 0.7^0\cdot 0.3^{10}\doteq 5.9049\cdot 10^{-6} \\ {C}_1(10)=\left(\genfrac{}{}{0ex}{}{10}{1}\right)=\frac{10!}{1!(10-1)!}=\frac{10}{1}=10 \\ {b}_1=\left(\genfrac{}{}{0ex}{}{10}{1}\right)\cdot k^1\cdot q^{10-1}=10\cdot 0.7^1\cdot 0.3^{10-1}\doteq 0.0001 \\ {C}_2(10)=\left(\genfrac{}{}{0ex}{}{10}{2}\right)=\frac{10!}{2!(10-2)!}=\frac{10\cdot 9}{2\cdot 1}=45 \\ {b}_2=\left(\genfrac{}{}{0ex}{}{10}{2}\right)\cdot k^2\cdot q^{10-2}=45\cdot 0.7^2\cdot 0.3^{10-2}\doteq 0.0014 \\ {C}_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ {b}_3=\left(\genfrac{}{}{0ex}{}{10}{3}\right)\cdot k^3\cdot q^{10-3}=120\cdot 0.7^3\cdot 0.3^{10-3}\doteq 0.009 \\ p=1-(b_0+b_1+b_2+b_3)=1-(5.9049\cdot 10^{-6}+0.0001+0.0014+0.009)=0.9894 \end{gathered}$$

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