Question: The probability that a new airport will get an award for its design is $0.16$, the probability that it will get an award for the efficient use of materials is $0.24$, and the probability that it will get both awards is $0.11$. Find the probability that (a) it will get at least one of the two awards, (b) it will get only one of the two awards, (c) it will get none of the two awards.

1. **State the problem:** We are given probabilities for two events: getting an award for design ($P(D) = 0.16$), getting an award for efficient use of materials ($P(M) = 0.24$), and getting both awards ($P(D \cap M) = 0.11$). We need to find: (a) Probability of getting at least one award: $P(D \cup M)$ (b) Probability of getting only one award (c) Probability of getting none of the awards 2. **Recall the formula for union of two events:** $$P(D \cup M) = P(D) + P(M) - P(D \cap M)$$ 3. **Calculate (a) probability of at least one award:** $$P(D \cup M) = 0.16 + 0.24 - 0.11 = 0.40 - 0.11 = 0.29$$ 4. **Calculate (b) probability of only one award:** Only one award means either design or materials but not both. $$P(\text{only one}) = P(D \cup M) - P(D \cap M) = 0.29 - 0.11 = 0.18$$ Alternatively, use: $$P(\text{only one}) = P(D) + P(M) - 2P(D \cap M) = 0.16 + 0.24 - 2(0.11) = 0.40 - 0.22 = 0.18$$ 5. **Calculate (c) probability of none of the awards:** Probability of none is the complement of getting at least one: $$P(\text{none}) = 1 - P(D \cup M) = 1 - 0.29 = 0.71$$ **Final answers:** (a) $0.29$ (b) $0.18$ (c) $0.71$