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,

The probability that four or fewer women are selected for 22 positions is,
![\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*10^(-7)+5.2452*10^(-6)+5.5075*10^(-5)+3.6716*10^(-4)+1.7440*10^(-3) \\ =0.00217175 \end{gathered}]()
Answer: P( at most four ) = 0.00217175 (nearest to 8 decimal places)