A survey was taken of a group's viewing habits of sporting events on TV during I.I-5 the last year. Let $A=\{$ watched football $\}, B=\{$ watched basketball $\}, C=\{$ watched baseball $\}$. The results indicate that if a person is selected at random from the surveyed group, then $P(A)=0.43, P(B)=0.40, P(C)=0.32, P(A \cap B)=0.29$, $P(A \cap C)=0.22, P(B \cap C)=0.20$, and $P(A \cap B \cap C)=0.15$. Find $P(A \cup B \cup C)$.
$$
\begin{aligned}
P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C) \\
& -P(B \cap C)+P(A \cap B \cap C) \\
= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \\
= & 0.59
\end{aligned}
$$