Answer:
[tex]$\text{P(Choose Cab Once)}=\frac{4}{9} $[/tex]
Step-by-step explanation:
In the morning Elizabeth has three transportation options to work.
It is [tex]$\frac{1}{3} $[/tex] :
In the evening Elizabeth has the same three transportation options to go home.
It is [tex]$\frac{1}{3} $[/tex] again.
But if we consider that she will take the cab once, the options will change to [tex]$\frac{2}{3} $[/tex] because now she can choose 2 options out of 3.
[tex]$\text{P(no cab) }= \frac{2}{3} $[/tex]
In order to use the cab once, we have two options:
She will use the cab in the morning or she will use the cab in the evening:
First route (cab in the morning):
[tex]$\text{Route 1}=\text{P(Choose Cab) } \cdot \text{P(Choose another one)} $[/tex]
[tex]$\frac{1}{3} \cdot \frac{2}{3} $[/tex]
Second route (cab in the evening):
[tex]$\text{Rote 2}=\text{P(Choose another one) } \cdot \text{P(Choose Cab)} $[/tex]
[tex]$\frac{1}{3} \cdot \frac{2}{3} $[/tex]
We have two routes, so
[tex]\text{P(Choose Cab Once)}=\text{Route 1}+\text{Route 2}[/tex]
[tex]$\text{P(Choose Cab Once)}=\frac{1}{3} \cdot \frac{2}{3} +\frac{1}{3} \cdot \frac{2}{3} $[/tex]
[tex]$\text{P(Choose Cab Once)}=\frac{2}{9} \cdot \frac{2}{9} $[/tex]
[tex]$\text{P(Choose Cab Once)}=\frac{4}{9} $[/tex]
