Let the probability that an eployee will be late to work be given by:
[tex]P(1|1)=0.21[/tex]Then, in a department of 5 employes, the problability that 1 or 2 employes are late is given by:
[tex]\begin{gathered} P((1\lor2)|5)=P(1|5)+P(2|5) \\ P(1|5)=P(1|1)\cdot(1-P(1|1))^4\cdot\frac{5!}{1!\cdot4!}=0.21\cdot0.79^4\cdot5=0.41 \\ P(2|5)=P(1|1)^2\cdot)(1-P(1|1))^3\cdot\frac{5!}{2!\cdot3!}=0.21^2\cdot0.79^3\cdot10=0.22 \\ \therefore P((1\lor2)|5)=0.41+0.22=0.63 \end{gathered}[/tex]