## Elementary Statistics (12th Edition)

We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, and also $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Let A=true positive, B=true negative, then $P(A\cup B)=\frac{119}{300}+\frac{154}{300}-\frac{0}{300}=\frac{273}{300}=0.91.$, which is the complement of the result from Exercise 37.