Manufacturer 24801

Five hundred of the products in the series are to be inspected with a repeat check. The manufacturer guarantees 2% scrap for a given production. Determine the probability of scraps among the 500 products reviewed between 12 and 20.

Correct answer:

p =  0.2051

Step-by-step explanation:

$$\begin{gathered} C_{12}(500)=\left(\genfrac{}{}{0ex}{}{500}{12}\right)=\frac{500!}{12!(500-12)!}\approx 4.462\times 10^{23}=446202084718341864844750 \\ {C}_{13}(500)=\left(\genfrac{}{}{0ex}{}{500}{13}\right)=\frac{500!}{13!(500-13)!}\approx 1.674\times 10^{25}=16749739795580833080326000 \\ {C}_{14}(500)=\left(\genfrac{}{}{0ex}{}{500}{14}\right)=\frac{500!}{14!(500-14)!}\approx 5.826\times 10^{26}=582651662889133265008483000 \\ {C}_{15}(500)=\left(\genfrac{}{}{0ex}{}{500}{15}\right)=\frac{500!}{15!(500-15)!}\approx 1.887\times 10^{28}=18877913877607917786274849200 \\ {C}_{16}(500)=\left(\genfrac{}{}{0ex}{}{500}{16}\right)=\frac{500!}{16!(500-16)!}\approx 5.722\times 10^{29}=572236764414990007896456366375 \\ {C}_{17}(500)=\left(\genfrac{}{}{0ex}{}{500}{17}\right)=\frac{500!}{17!(500-17)!}\approx 1.629\times 10^{31}=16291917292756186107169698901500 \\ {C}_{18}(500)=\left(\genfrac{}{}{0ex}{}{500}{18}\right)=\frac{500!}{18!(500-18)!}\approx 4.371\times 10^{32}=437166447355624327209053587190250 \\ {C}_{19}(500)=\left(\genfrac{}{}{0ex}{}{500}{19}\right)=\frac{500!}{19!(500-19)!}\approx 1.109\times 10^{34}=11090222506600575037619148896089500 \\ {C}_{20}(500)=\left(\genfrac{}{}{0ex}{}{500}{20}\right)=\frac{500!}{20!(500-20)!}\approx 2.667\times 10^{35}=266719851283743829654740530950952475 \\ n=500 \\ q=2\%=\frac{2}{100}=\frac{1}{50}=0.02 \\ {p}_{12}=\left(\genfrac{}{}{0ex}{}{n}{12}\right)\cdot q^{12}\cdot (1-q{)}^{n-12}=446202084718341864844750\cdot 0.02^{12}\cdot (1-0.02{)}^{500-12}\doteq 0.0955 \\ {p}_{13}=\left(\genfrac{}{}{0ex}{}{n}{13}\right)\cdot q^{13}\cdot (1-q{)}^{n-13}=16749739795580833080326000\cdot 0.02^{13}\cdot (1-0.02{)}^{500-13}\doteq 0.0732 \\ {p}_{14}=\left(\genfrac{}{}{0ex}{}{n}{14}\right)\cdot q^{14}\cdot (1-q{)}^{n-14}=582651662889133265008483000\cdot 0.02^{14}\cdot (1-0.02{)}^{500-14}\doteq 0.052 \\ {p}_{15}=\left(\genfrac{}{}{0ex}{}{n}{15}\right)\cdot q^{15}\cdot (1-q{)}^{n-15}=18877913877607917786274849200\cdot 0.02^{15}\cdot (1-0.02{)}^{500-15}\doteq 0.0344 \\ {p}_{16}=\left(\genfrac{}{}{0ex}{}{n}{16}\right)\cdot q^{16}\cdot (1-q{)}^{n-16}=572236764414990007896456366375\cdot 0.02^{16}\cdot (1-0.02{)}^{500-16}\doteq 0.0213 \\ {p}_{17}=\left(\genfrac{}{}{0ex}{}{n}{17}\right)\cdot q^{17}\cdot (1-q{)}^{n-17}=16291917292756186107169698901500\cdot 0.02^{17}\cdot (1-0.02{)}^{500-17}\doteq 0.0124 \\ {p}_{18}=\left(\genfrac{}{}{0ex}{}{n}{18}\right)\cdot q^{18}\cdot (1-q{)}^{n-18}=437166447355624327209053587190250\cdot 0.02^{18}\cdot (1-0.02{)}^{500-18}\doteq 0.0068 \\ {p}_{19}=\left(\genfrac{}{}{0ex}{}{n}{19}\right)\cdot q^{19}\cdot (1-q{)}^{n-19}=11090222506600575037619148896089500\cdot 0.02^{19}\cdot (1-0.02{)}^{500-19}\doteq 0.0035 \\ {p}_{20}=\left(\genfrac{}{}{0ex}{}{n}{20}\right)\cdot q^{20}\cdot (1-q{)}^{n-20}=266719851283743829654740530950952475\cdot 0.02^{20}\cdot (1-0.02{)}^{500-20}\doteq 0.0017 \\ p=p_{13}+p_{14}+p_{15}+p_{16}+p_{17}+p_{18}+p_{19}+p_{20}=0.0732+0.052+0.0344+0.0213+0.0124+0.0068+0.0035+0.0017=0.2051 \end{gathered}$$

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