Alopecia

Medical literature indicates that 45% of men suffer from alopecia. For random sample of 8 men, calculate the probability that:
(a) exactly four men suffer from alopecia.
(b) at most two men suffer from alopecia.

Correct result:

p₄ =  0.2627
p =  0.2201

Solution:

$$\begin{gathered} C_4(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{4}}\right)={\displaystyle \frac{8!}{4!(8-4)!}}={\displaystyle \frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 1}}=70 \\ q=45\mathrm{\%}={\displaystyle \frac{45}{100}}=0.45 \\ n=8 \\ {p}_4=\left(\genfrac{}{}{0px}{}{n}{4}\right)\cdot q^4\cdot (1-q{)}^{n-4}=70\cdot 0.45^4\cdot (1-0.45{)}^{8-4}=0.2627 \end{gathered}$$

$$\begin{gathered} C_0(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{0}}\right)={\displaystyle \frac{8!}{0!(8-0)!}}={\displaystyle \frac{1}{1}}=1 \\ {C}_1(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{1}}\right)={\displaystyle \frac{8!}{1!(8-1)!}}={\displaystyle \frac{8}{1}}=8 \\ {C}_2(8)=\left({\displaystyle \genfrac{}{}{0px}{}{8}{2}}\right)={\displaystyle \frac{8!}{2!(8-2)!}}={\displaystyle \frac{8\cdot 7}{2\cdot 1}}=28 \\ {p}_0=\left(\genfrac{}{}{0px}{}{n}{0}\right)\cdot q^0\cdot (1-q{)}^{n-0}=1\cdot 0.45^0\cdot (1-0.45{)}^{8-0}\doteq 0.0084 \\ {p}_1=\left(\genfrac{}{}{0px}{}{n}{1}\right)\cdot q^1\cdot (1-q{)}^{n-1}=8\cdot 0.45^1\cdot (1-0.45{)}^{8-1}\doteq 0.0548 \\ {p}_2=\left(\genfrac{}{}{0px}{}{n}{2}\right)\cdot q^2\cdot (1-q{)}^{n-2}=28\cdot 0.45^2\cdot (1-0.45{)}^{8-2}\doteq 0.1569 \\ p=p_0+p_1+p_2=0.0084+0.0548+0.1569=0.2201 \end{gathered}$$

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