Flu probability

In the monitored group of people, 8% are ill with the flu. One hundred people from this group were examined. What is the probability that at most 5 of them will be sick with the flu?
(round to 3 decimal places)

Final Answer:

p =  17.9876 %

Step-by-step explanation:

$$\begin{gathered} q=8\%=\frac{8}{100}=\frac{2}{25}=0.08 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{100}{0}\right)\cdot q^0\cdot (1-q{)}^{100-0}=(\genfrac{}{}{0ex}{}{100}{0})\cdot 0.08^0\cdot (1-0.08{)}^{100-0}=1\cdot 0.08^0\cdot 0.92^{100}\doteq 0.0002 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{100}{1}\right)\cdot q^1\cdot (1-q{)}^{100-1}=(\genfrac{}{}{0ex}{}{100}{1})\cdot 0.08^1\cdot (1-0.08{)}^{100-1}=100\cdot 0.08^1\cdot 0.92^{99}\doteq 0.0021 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{100}{2}\right)\cdot q^2\cdot (1-q{)}^{100-2}=(\genfrac{}{}{0ex}{}{100}{2})\cdot 0.08^2\cdot (1-0.08{)}^{100-2}=4950\cdot 0.08^2\cdot 0.92^{98}\doteq 0.009 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{100}{3}\right)\cdot q^3\cdot (1-q{)}^{100-3}=(\genfrac{}{}{0ex}{}{100}{3})\cdot 0.08^3\cdot (1-0.08{)}^{100-3}=161700\cdot 0.08^3\cdot 0.92^{97}\doteq 0.0254 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{100}{4}\right)\cdot q^4\cdot (1-q{)}^{100-4}=(\genfrac{}{}{0ex}{}{100}{4})\cdot 0.08^4\cdot (1-0.08{)}^{100-4}=3921225\cdot 0.08^4\cdot 0.92^{96}\doteq 0.0536 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{100}{5}\right)\cdot q^5\cdot (1-q{)}^{100-5}=(\genfrac{}{}{0ex}{}{100}{5})\cdot 0.08^5\cdot (1-0.08{)}^{100-5}=75287520\cdot 0.08^5\cdot 0.92^{95}\doteq 0.0895 \\ p=100\cdot (p_0+p_1+p_2+p_3+p_4+p_5)=100\cdot (0.0002+0.0021+0.009+0.0254+0.0536+0.0895)=17.9876\% \end{gathered}$$

Try another example