Given:
The number of employee=22
The probability of selecting women is 1/2. As there are equal number of qualified men as qualified women.
The data follows binomial distribution,
[tex]\begin{gathered} X\text{ \textasciitilde{}B(n,p,q)} \\ n=22 \\ p=\frac{1}{2}=0.5 \\ q=1-p=1-0.5=0.5 \\ P(X=x)=^nC_xp^xq^{n-x} \end{gathered}[/tex]The probability that four or fewer women are selected for 22 positions is,
[tex]\begin{gathered} P(X\leq4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =^{22}C_0(0.5)^0(0.5)^{22-0}+^{22}C_1(0.5)^1(0.5)^{22-1}+^{22}C_2(0.5)^2(0.5)^{22-2}+^{22}C_3(0.5)^3(0.5)^{22-3}+^{22}C_4(0.5)^4(0.5)^{22-4} \\ =2.3841\times10^{-7}+5.2452\times10^{-6}+5.5075\times10^{-5}+3.6716\times10^{-4}+1.7440\times10^{-3} \\ =0.00217175 \end{gathered}[/tex]Answer: P( at most four ) = 0.00217175 (nearest to 8 decimal places)