Solution :
Case I :
If Collen is late on [tex]0[/tex] out of [tex]5[/tex] days.
[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} $[/tex]
[tex]$=\frac{1}{32}[/tex]
Case II :
When Collen is late on [tex]1[/tex] out of [tex]5[/tex] days.
[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_1$[/tex]
[tex]$=\frac{1}{32} \times 5$[/tex]
[tex]$=\frac{5}{32}[/tex]
Case III :
When Collen was late on [tex]2[/tex] out of [tex]5[/tex] days.
[tex]$= \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times ^5C_2$[/tex]
[tex]$=\frac{1}{32} \times 10$[/tex]
[tex]$=\frac{5}{16}[/tex]
Therefore, the [tex]\text{probability}[/tex] that Collen will arrive late to work no more than [tex]\text{twice}[/tex] during a [tex]\text{five day workweek}[/tex] is :
[tex]$=\frac{1}{32} + \frac{5}{32} + \frac{5}{16} $[/tex]
[tex]$=\frac{1}{2}$[/tex]