Deficiencies

The hygienic inspection of 2000 mass caterers found deficiencies in 300 establishments. What is the probability that flaws in a maximum of 3 devices will be found during the inspection of 10 devices?

Final Answer:

p =  95.003 %

Step-by-step explanation:

$$\begin{gathered} C_0(10)=\left(\genfrac{}{}{0ex}{}{10}{0}\right)=\frac{10!}{0!(10-0)!}=\frac{1}{1}=1 \\ {C}_1(10)=\left(\genfrac{}{}{0ex}{}{10}{1}\right)=\frac{10!}{1!(10-1)!}=\frac{10}{1}=10 \\ {C}_2(10)=\left(\genfrac{}{}{0ex}{}{10}{2}\right)=\frac{10!}{2!(10-2)!}=\frac{10\cdot 9}{2\cdot 1}=45 \\ {C}_3(10)=\left(\genfrac{}{}{0ex}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120 \\ q=\frac{300}{2000}=\frac{3}{20}=0.15 \\ n=10 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot q^0\cdot (1-q{)}^{n-0}=1\cdot 0.15^0\cdot (1-0.15{)}^{10-0}\doteq 0.1969 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot q^1\cdot (1-q{)}^{n-1}=10\cdot 0.15^1\cdot (1-0.15{)}^{10-1}\doteq 0.3474 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot q^2\cdot (1-q{)}^{n-2}=45\cdot 0.15^2\cdot (1-0.15{)}^{10-2}\doteq 0.2759 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot q^3\cdot (1-q{)}^{n-3}=120\cdot 0.15^3\cdot (1-0.15{)}^{10-3}\doteq 0.1298 \\ p=100\cdot (p_0+p_1+p_2+p_3)=100\cdot (0.1969+0.3474+0.2759+0.1298)=95.003\% \end{gathered}$$

Try another example