Doctors

The drug successfully treats 90% of cases. Calculate the probability that he will cure at least 18 patients out of 20?

Correct answer:

P =  67.6927 %

Step-by-step explanation:

$$\begin{gathered} p=90\mathrm{\%}={\displaystyle \frac{90}{100}}=0.9 \\ {C}_{18}(20)=\left({\displaystyle \genfrac{}{}{0px}{}{20}{18}}\right)={\displaystyle \frac{20!}{18!(20-18)!}}={\displaystyle \frac{20\cdot 19}{2\cdot 1}}=190 \\ {p}_{18}=\left(\genfrac{}{}{0px}{}{20}{18}\right)\cdot p^{18}\cdot (1-p{)}^2=190\cdot 0.9^{18}\cdot (1-0.9{)}^2\doteq 0.2852 \\ {C}_{19}(20)=\left({\displaystyle \genfrac{}{}{0px}{}{20}{19}}\right)={\displaystyle \frac{20!}{19!(20-19)!}}={\displaystyle \frac{20}{1}}=20 \\ {p}_{19}=\left(\genfrac{}{}{0px}{}{20}{19}\right)\cdot p^{19}\cdot (1-p{)}^1=20\cdot 0.9^{19}\cdot (1-0.9{)}^1\doteq 0.2702 \\ {C}_{20}(20)=\left({\displaystyle \genfrac{}{}{0px}{}{20}{20}}\right)={\displaystyle \frac{20!}{20!(20-20)!}}={\displaystyle \frac{1}{1}}=1 \\ {p}_{20}=\left(\genfrac{}{}{0px}{}{20}{20}\right)\cdot p^{20}\cdot (1-p{)}^0=1\cdot 0.9^{20}\cdot (1-0.9{)}^0\doteq 0.1216 \\ P=100\cdot (p_{18}+p_{19}+p_{20})=100\cdot (0.2852+0.2702+0.1216)=67.6927\mathrm{\%} \end{gathered}$$

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