Probability 7627

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)

Correct answer:

p =  17.9876 %

Step-by-step explanation:

$$\begin{gathered} q=8/100=\frac{2}{25}=0.08 \\ w=1-q=1-0.08=\frac{23}{25}=0.92 \\ {C}_0(100)=\left(\genfrac{}{}{0ex}{}{100}{0}\right)=\frac{100!}{0!(100-0)!}=\frac{1}{1}=1 \\ {p}_0=\left(\genfrac{}{}{0ex}{}{100}{0}\right)\cdot q^0\cdot w^{100}=1\cdot 0.08^0\cdot 0.92^{100}\doteq 0.0002 \\ {C}_1(100)=\left(\genfrac{}{}{0ex}{}{100}{1}\right)=\frac{100!}{1!(100-1)!}=\frac{100}{1}=100 \\ {p}_1=\left(\genfrac{}{}{0ex}{}{100}{1}\right)\cdot q^1\cdot w^{99}=100\cdot 0.08^1\cdot 0.92^{99}\doteq 0.0021 \\ {C}_2(100)=\left(\genfrac{}{}{0ex}{}{100}{2}\right)=\frac{100!}{2!(100-2)!}=\frac{100\cdot 99}{2\cdot 1}=4950 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{100}{2}\right)\cdot q^2\cdot w^{98}=4950\cdot 0.08^2\cdot 0.92^{98}\doteq 0.009 \\ {C}_3(100)=\left(\genfrac{}{}{0ex}{}{100}{3}\right)=\frac{100!}{3!(100-3)!}=\frac{100\cdot 99\cdot 98}{3\cdot 2\cdot 1}=161700 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{100}{3}\right)\cdot q^3\cdot w^{97}=161700\cdot 0.08^3\cdot 0.92^{97}\doteq 0.0254 \\ {C}_4(100)=\left(\genfrac{}{}{0ex}{}{100}{4}\right)=\frac{100!}{4!(100-4)!}=\frac{100\cdot 99\cdot 98\cdot 97}{4\cdot 3\cdot 2\cdot 1}=3921225 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{100}{4}\right)\cdot q^4\cdot w^{96}=3921225\cdot 0.08^4\cdot 0.92^{96}\doteq 0.0536 \\ {C}_5(100)=\left(\genfrac{}{}{0ex}{}{100}{5}\right)=\frac{100!}{5!(100-5)!}=\frac{100\cdot 99\cdot 98\cdot 97\cdot 96}{5\cdot 4\cdot 3\cdot 2\cdot 1}=75287520 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{100}{5}\right)\cdot q^5\cdot w^{95}=75287520\cdot 0.08^5\cdot 0.92^{95}\doteq 0.0895 \\ p=100\cdot (p_0+p_1+p_2+p_3+p_4+p_5)=100\cdot (0.0002+0.0021+0.009+0.0254+0.0536+0.0895)=17.9876\% \end{gathered}$$

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