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 \(\displaystyle{P}{\left({A}\right)}={0}+{0.42}\cdot{0.7}={0.294}\)