Repairs

A small repair shop orders parts for its repairs. Its supplier claims that only 2% of its parts are defective. In order of 500 parts, an employee takes a sample of 100 parts. Determine the likelihood that the employee finds three or more parts are defective.

Final Answer:

p =  0.3233

Step-by-step explanation:

$$\begin{gathered} r=2\%=\frac{2}{100}=\frac{1}{50}=0.02 \\ n=500 \\ {n}_2=100 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n_2}{0}\right)\cdot r^0\cdot (1-r{)}^{n_2-0}=(\genfrac{}{}{0ex}{}{100}{0})\cdot 0.02^0\cdot (1-0.02{)}^{100-0}=1\cdot 0.02^0\cdot 0.98^{100}\doteq 0.1326 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n_2}{1}\right)\cdot r^1\cdot (1-r{)}^{n_2-1}=(\genfrac{}{}{0ex}{}{100}{1})\cdot 0.02^1\cdot (1-0.02{)}^{100-1}=100\cdot 0.02^1\cdot 0.98^{99}\doteq 0.2707 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n_2}{2}\right)\cdot r^2\cdot (1-r{)}^{n_2-2}=(\genfrac{}{}{0ex}{}{100}{2})\cdot 0.02^2\cdot (1-0.02{)}^{100-2}=4950\cdot 0.02^2\cdot 0.98^{98}\doteq 0.2734 \\ P=p_0+p_1+p_2=0.1326+0.2707+0.2734\doteq 0.6767 \\ p=1-P=1-0.6767=0.3233 \end{gathered}$$

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