An experiment has three events $A$, $B$, and $C$ with $P(A) = 0.23$, $P(B) = 0.40$, and $P(C) = 0.85$. Events $A$ and $B$ are **mutually exclusive**, and events $B$ and $C$ are **independent**. What is $P(B \text{ or } C)$?

Apply the **inclusion-exclusion** rule for *or*: $P(B \text{ or } C) = P(B) + P(C) - P(B \text{ and } C)$. Since $B$ and $C$ are **independent**, $P(B \text{ and } C) = P(B) \cdot P(C) = (0.40)(0.85) = 0.34$. Therefore $P(B \text{ or } C) = 0.40 + 0.85 - 0.34 = 1.25 - 0.34 = 0.91$. Why the others fail: - **A** ($0.34$) is $P(B \text{ and } C)$ — the intersection, not the union. - **B** ($0.63 = 0.23 + 0.40$) is $P(A \text{ or } B)$ — using the wrong pair of events. - **D** ($1.25 = 0.40 + 0.85$) forgets to subtract the overlap. Probabilities cannot exceed 1, which is an immediate red flag. Note that *mutually exclusive* and *independent* are **different** properties — if $P(B), P(C) \ne 0$, the same pair cannot be both. The problem applies one property to one pair and the other to a different pair.