Drunken drivers

40% of drivers driving between 11 pm and 5 am are drunken drivers. In a random sample of 20 drivers driving between 11 pm and 5 am, find the probability that:
A) Exactly 12 will be drunken drivers
B) At least 7 will be drunken drivers
C) At most 5 will be drunken drivers

Correct answer:

p₁₂ =  0.0355
b =  0.75
c =  0.1256

Step-by-step explanation:

$$\begin{gathered} p=40\%=\frac{40}{100}=\frac{2}{5}=0.4 \\ n=20 \\ {p}_{12}=\left(\genfrac{}{}{0ex}{}{n}{12}\right)\cdot p^{12}\cdot (1-p{)}^{n-12}=(\genfrac{}{}{0ex}{}{20}{12})\cdot 0.4^{12}\cdot (1-0.4{)}^{20-12}=125970\cdot 0.4^{12}\cdot 0.6^8=0.0355 \end{gathered}$$

$$\begin{gathered} p_0=\left(\genfrac{}{}{0ex}{}{n}{0}\right)\cdot p^0\cdot (1-p{)}^{n-0}=(\genfrac{}{}{0ex}{}{20}{0})\cdot 0.4^0\cdot (1-0.4{)}^{20-0}=1\cdot 0.4^0\cdot 0.6^{20}\doteq 3.6562\cdot 10^{-5} \\ {p}_1=\left(\genfrac{}{}{0ex}{}{n}{1}\right)\cdot p^1\cdot (1-p{)}^{n-1}=(\genfrac{}{}{0ex}{}{20}{1})\cdot 0.4^1\cdot (1-0.4{)}^{20-1}=20\cdot 0.4^1\cdot 0.6^{19}\doteq 0.0005 \\ {p}_2=\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot p^2\cdot (1-p{)}^{n-2}=(\genfrac{}{}{0ex}{}{20}{2})\cdot 0.4^2\cdot (1-0.4{)}^{20-2}=190\cdot 0.4^2\cdot 0.6^{18}\doteq 0.0031 \\ {p}_3=\left(\genfrac{}{}{0ex}{}{n}{3}\right)\cdot p^3\cdot (1-p{)}^{n-3}=(\genfrac{}{}{0ex}{}{20}{3})\cdot 0.4^3\cdot (1-0.4{)}^{20-3}=1140\cdot 0.4^3\cdot 0.6^{17}\doteq 0.0123 \\ {p}_4=\left(\genfrac{}{}{0ex}{}{n}{4}\right)\cdot p^4\cdot (1-p{)}^{n-4}=(\genfrac{}{}{0ex}{}{20}{4})\cdot 0.4^4\cdot (1-0.4{)}^{20-4}=4845\cdot 0.4^4\cdot 0.6^{16}\doteq 0.035 \\ {p}_5=\left(\genfrac{}{}{0ex}{}{n}{5}\right)\cdot p^5\cdot (1-p{)}^{n-5}=(\genfrac{}{}{0ex}{}{20}{5})\cdot 0.4^5\cdot (1-0.4{)}^{20-5}=15504\cdot 0.4^5\cdot 0.6^{15}\doteq 0.0746 \\ {p}_6=\left(\genfrac{}{}{0ex}{}{n}{6}\right)\cdot p^6\cdot (1-p{)}^{n-6}=(\genfrac{}{}{0ex}{}{20}{6})\cdot 0.4^6\cdot (1-0.4{)}^{20-6}=38760\cdot 0.4^6\cdot 0.6^{14}\doteq 0.1244 \\ {p}_7=\left(\genfrac{}{}{0ex}{}{n}{7}\right)\cdot p^7\cdot (1-p{)}^{n-7}=(\genfrac{}{}{0ex}{}{20}{7})\cdot 0.4^7\cdot (1-0.4{)}^{20-7}=77520\cdot 0.4^7\cdot 0.6^{13}\doteq 0.1659 \\ c=p_0+p_1+p_2+p_3+p_4+p_5=3.6562\cdot 10^{-5}+0.0005+0.0031+0.0123+0.035+0.0746\doteq 0.1256 \\ b=1-(c+p_6)=1-(0.1256+0.1244)=0.75 \end{gathered}$$

Try another example