A disease

A disease affects 10% of the individuals in a population, and a sample of 100 people was selected from the population. What is the probability of finding the disease in at least 15 people?

Correct answer:

p =  0.0726

Step-by-step explanation:

$$\begin{gathered} r=10\%=\frac{10}{100}=\frac{1}{10}=0.1 \\ n=100 \\ {n}_1=15 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot r^0\cdot (1-r{)}^{n-0}=(\genfrac{}{}{0ex}{}{100}{0})\cdot 0.1^0\cdot (1-0.1{)}^{100-0}=1\cdot 0.1^0\cdot 0.9^{100}\doteq 2.6561\cdot 10^{-5} \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot r^1\cdot (1-r{)}^{n-1}=(\genfrac{}{}{0ex}{}{100}{1})\cdot 0.1^1\cdot (1-0.1{)}^{100-1}=100\cdot 0.1^1\cdot 0.9^{99}\doteq 0.0003 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot r^2\cdot (1-r{)}^{n-2}=(\genfrac{}{}{0ex}{}{100}{2})\cdot 0.1^2\cdot (1-0.1{)}^{100-2}=4950\cdot 0.1^2\cdot 0.9^{98}\doteq 0.0016 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot r^3\cdot (1-r{)}^{n-3}=(\genfrac{}{}{0ex}{}{100}{3})\cdot 0.1^3\cdot (1-0.1{)}^{100-3}=161700\cdot 0.1^3\cdot 0.9^{97}\doteq 0.0059 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{n}{4}\right)\cdot r^4\cdot (1-r{)}^{n-4}=(\genfrac{}{}{0ex}{}{100}{4})\cdot 0.1^4\cdot (1-0.1{)}^{100-4}=3921225\cdot 0.1^4\cdot 0.9^{96}\doteq 0.0159 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{n}{5}\right)\cdot r^5\cdot (1-r{)}^{n-5}=(\genfrac{}{}{0ex}{}{100}{5})\cdot 0.1^5\cdot (1-0.1{)}^{100-5}=75287520\cdot 0.1^5\cdot 0.9^{95}\doteq 0.0339 \\ {p}_6=\left(\genfrac{}{}{0ex}{}{n}{6}\right)\cdot r^6\cdot (1-r{)}^{n-6}=(\genfrac{}{}{0ex}{}{100}{6})\cdot 0.1^6\cdot (1-0.1{)}^{100-6}=1192052400\cdot 0.1^6\cdot 0.9^{94}\doteq 0.0596 \\ {p}_7=\left(\genfrac{}{}{0ex}{}{n}{7}\right)\cdot r^7\cdot (1-r{)}^{n-7}=(\genfrac{}{}{0ex}{}{100}{7})\cdot 0.1^7\cdot (1-0.1{)}^{100-7}=16007560800\cdot 0.1^7\cdot 0.9^{93}\doteq 0.0889 \\ {p}_8=\left(\genfrac{}{}{0ex}{}{n}{8}\right)\cdot r^8\cdot (1-r{)}^{n-8}=(\genfrac{}{}{0ex}{}{100}{8})\cdot 0.1^8\cdot (1-0.1{)}^{100-8}=186087894300\cdot 0.1^8\cdot 0.9^{92}\doteq 0.1148 \\ {p}_9=\left(\genfrac{}{}{0ex}{}{n}{9}\right)\cdot r^9\cdot (1-r{)}^{n-9}=(\genfrac{}{}{0ex}{}{100}{9})\cdot 0.1^9\cdot (1-0.1{)}^{100-9}=1902231808400\cdot 0.1^9\cdot 0.9^{91}\doteq 0.1304 \\ {p}_{10}=\left(\genfrac{}{}{0ex}{}{n}{10}\right)\cdot r^{10}\cdot (1-r{)}^{n-10}=(\genfrac{}{}{0ex}{}{100}{10})\cdot 0.1^{10}\cdot (1-0.1{)}^{100-10}=17310309456440\cdot 0.1^{10}\cdot 0.9^{90}\doteq 0.1319 \\ {p}_{11}=\left(\genfrac{}{}{0ex}{}{n}{11}\right)\cdot r^{11}\cdot (1-r{)}^{n-11}=(\genfrac{}{}{0ex}{}{100}{11})\cdot 0.1^{11}\cdot (1-0.1{)}^{100-11}=141629804643600\cdot 0.1^{11}\cdot 0.9^{89}\doteq 0.1199 \\ {p}_{12}=\left(\genfrac{}{}{0ex}{}{n}{12}\right)\cdot r^{12}\cdot (1-r{)}^{n-12}=(\genfrac{}{}{0ex}{}{100}{12})\cdot 0.1^{12}\cdot (1-0.1{)}^{100-12}=1050421051106700\cdot 0.1^{12}\cdot 0.9^{88}\doteq 0.0988 \\ {p}_{13}=\left(\genfrac{}{}{0ex}{}{n}{13}\right)\cdot r^{13}\cdot (1-r{)}^{n-13}=(\genfrac{}{}{0ex}{}{100}{13})\cdot 0.1^{13}\cdot (1-0.1{)}^{100-13}=7110542499799200\cdot 0.1^{13}\cdot 0.9^{87}\doteq 0.0743 \\ {p}_{14}=\left(\genfrac{}{}{0ex}{}{n}{14}\right)\cdot r^{14}\cdot (1-r{)}^{n-14}=(\genfrac{}{}{0ex}{}{100}{14})\cdot 0.1^{14}\cdot (1-0.1{)}^{100-14}=44186942677323600\cdot 0.1^{14}\cdot 0.9^{86}\doteq 0.0513 \\ a=p_0+p_1+p_2+p_3+p_4+p_5+p_6+p_7+p_8+p_9+p_{10}+p_{11}+p_{12}+p_{13}+p_{14}=2.6561\cdot 10^{-5}+0.0003+0.0016+0.0059+0.0159+0.0339+0.0596+0.0889+0.1148+0.1304+0.1319+0.1199+0.0988+0.0743+0.0513\doteq 0.9274 \\ p=1-a=1-0.9274=0.0726 \end{gathered}$$

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