1. **State the problem:** We are given the probabilities of intersections of three types of trees (Cedar, Cypress, Redwood) in quarter-acre plots. 2. **Given data:** - $P(C \cap Cy) = 0.40$ - $P(C \cap R) = 0.25$ - $P(Cy \cap R) = 0.20$ - $P(C \cap Cy \cap R) = 0.15$ 3. **Goal:** Find the probability that a randomly selected plot has exactly two of the three types of trees. 4. **Formula:** The probability of exactly two types is the sum of the probabilities of each pair intersection minus the triple intersection (since it is counted in all pairs): $$P(\text{exactly two}) = P(C \cap Cy) + P(C \cap R) + P(Cy \cap R) - 3 \times P(C \cap Cy \cap R)$$ 5. **Explanation:** The triple intersection is included in each pair intersection, so we subtract it three times to remove overcounting. 6. **Calculate:** $$P(\text{exactly two}) = 0.40 + 0.25 + 0.20 - 3 \times 0.15$$ $$= 0.85 - 0.45$$ $$= 0.40$$ 7. **Answer:** The probability that a randomly selected plot has exactly two of the three types of trees is **0.40** or 40%.