Given:
Number of employees are 5.
Probability that an employee will be late to work at a large corporation is 0.21.
The given data follows binomial distribution,
[tex]\begin{gathered} X\text{ \textasciitilde{}B(n,p,q)} \\ n=5 \\ p=0.21 \\ q=1-p=1-0.21=0.79 \\ P(X=x)=^nC_xp^xq^{n-x} \end{gathered}[/tex]The probability that at least 3 employees are late is given as,
[tex]\begin{gathered} P(X\ge3)=P(X=3)+P(X=4_{})+P(X=5) \\ P(X\ge3)=^5C_3(0.21)^3(0.79)^{5-3}+^5C_4(0.21)^4(0.79)^{5-4}+^5C_5(0.21)^5(0.79)^{5-5} \\ P(X\ge3)=0.0578+0.0077+0.0004 \\ P(X\ge3)=0.0659 \end{gathered}[/tex]Answer: The probability that in a department of 5 employees at least 3 are late is 0.0659.
