c. To show that P(A∪B)=P(A)+P(B)−P(A∩B) :
\begin{align*} P(A \cup B)&= 0.67\\ P(A) + P(B) -P(A \cap B) &=0.40 + 0.37 -0.10\\ &=0.67 \end{align*}P(A∪B)P(A)+P(B)−P(A∩B)=0.67=0.40+0.37−0.10=0.67
Answer
Use the values obtained in Exercise 4.56 to show that the equalities hold.