Solution
otal number of cCustomer = 11 + 9 + 11 + 5 + 5 + 1 + 3 = 45
Let A denotes the event that the Customer waited for at least 12 minutes
Let B denotes the event that the Customer waited between 8 and 15 minutes
[tex]\begin{gathered} n(A)=5+5+1+3=14 \\ n(B)=11+5=16 \\ n(A\cap B)=5 \end{gathered}[/tex]
We want to find
[tex]p(A\cup B)[/tex][tex]\begin{gathered} p(A\cup B)=p(A)+p(B)-p(A\cap B) \\ p(A\cup B)=\frac{14}{45}+\frac{16}{45}-\frac{5}{45} \\ p(A\cup B)=\frac{25}{45} \\ p(A\cup B)=\frac{5}{9} \\ p(A\cup B)=0.556\text{ (to the nearest thousandth)} \end{gathered}[/tex]
hus, the probability is 0.556
Option D
