1. **Problem statement:** We have 3 X-ray machines, each can be broken or not. Let $A_i$ be the event that machine $i$ is broken. We want to express the following events using $A_1, A_2, A_3$. 2. **Recall:** The event $A_i$ means machine $i$ is broken, and $\overline{A_i}$ means machine $i$ is not broken. 3. **a. Exactly 1 machine is broken:** This means one of the machines is broken and the other two are not. So the event is $$ (A_1 \cap \overline{A_2} \cap \overline{A_3}) \cup (\overline{A_1} \cap A_2 \cap \overline{A_3}) \cup (\overline{A_1} \cap \overline{A_2} \cap A_3) $$ 4. **b. Exactly 2 machines are broken:** This means two machines are broken and one is not. So the event is $$ (A_1 \cap A_2 \cap \overline{A_3}) \cup (A_1 \cap \overline{A_2} \cap A_3) \cup (\overline{A_1} \cap A_2 \cap A_3) $$ 5. **c. All 3 machines are broken:** This means all three events happen simultaneously: $$ A_1 \cap A_2 \cap A_3 $$ 6. **d. At least 1 machine is broken:** This means one or more machines are broken, which is the union of all three events: $$ A_1 \cup A_2 \cup A_3 $$ These expressions use basic set operations: intersection ($\cap$) means "and", union ($\cup$) means "or", and complement ($\overline{A_i}$) means "not". **Final answers:** - Exactly 1 broken: $ (A_1 \cap \overline{A_2} \cap \overline{A_3}) \cup (\overline{A_1} \cap A_2 \cap \overline{A_3}) \cup (\overline{A_1} \cap \overline{A_2} \cap A_3) $ - Exactly 2 broken: $ (A_1 \cap A_2 \cap \overline{A_3}) \cup (A_1 \cap \overline{A_2} \cap A_3) \cup (\overline{A_1} \cap A_2 \cap A_3) $ - All 3 broken: $ A_1 \cap A_2 \cap A_3 $ - At least 1 broken: $ A_1 \cup A_2 \cup A_3 $