# Geographical Analysis (Jan, 2010) presented a study of Emergency Medical Services (EMS) ability to meet the demand for an ambulance. In one example, t

Probability
Geographical Analysis (Jan, 2010) presented a study of Emergency Medical Services (EMS) ability to meet the demand for an ambulance. In one example, the researchers presented the following scenario. An ambulance station has one vehicle and two demand locations, A and B. The probability that the ambulance can travel to a location in under eight minutes is .58 for location A and .42 for location B. The probability that the ambulance is busy at any point in time is .3. a. Find the probability that EMS can meet demand for an ambulance at location A. b. Find the probability that EMS can meet demand for an ambulance at location B.

Let C denote the event that the ambulance is busy. Then $$P(C)=0.3$$ and $$\displaystyle{P}{\left({C}^{{c}}\right)}={0.7}$$. Let A denote the event that EMS can meet demand at location A, and let A denote the event that EMS can meet demand at location B.
We must find P(A), which we can write as $$\displaystyle{P}{\left({A}\right)}={P}{\left({A}{\mid}{C}\right)}{P}{\left({C}\right)}+{P}{\left({A}{\mid}{C}^{{c}}\right)}{P}{\left({C}^{{c}}\right)}$$
If C happened, then $$P(A|C)=0$$. On the other hand, $$\displaystyle{P}{\left({A}{\mid}{C}^{{c}}\right)}={0.58}$$, so $$\displaystyle{P}{\left({A}\right)}={0}+{0.58}\cdot{0.7}={0.406}$$
We must find P(B), which we can write as $$\displaystyle{P}{\left({B}\right)}={P}{\left({B}{\mid}{C}\right)}{P}{\left({C}\right)}+{P}{\left({B}{\mid}{C}^{{c}}\right)}{P}{\left({C}^{{c}}\right)}$$
If C happened, then $$P(B|C)=0$$. On the other hand, $$\displaystyle{P}{\left({B}{\mid}{C}^{{c}}\right)}={0.42}$$, so $$\displaystyle{P}{\left({A}\right)}={0}+{0.42}\cdot{0.7}={0.294}$$