1. **State the problem:** We have a survey of 1,000 university students. - 780 own laptops. - 470 own cars. - 310 own both cars and laptops. We want to find the empirical probabilities for: (A) Owning either a car or a laptop. (B) Owning neither a car nor a laptop. 2. **Formula used:** For two events $A$ and $B$, the probability of $A$ or $B$ is given by the formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ This accounts for the overlap so we don't double count students who own both. 3. **Calculate probabilities:** - Total students $= 1000$ - $P(\text{laptop}) = \frac{780}{1000} = 0.78$ - $P(\text{car}) = \frac{470}{1000} = 0.47$ - $P(\text{both}) = \frac{310}{1000} = 0.31$ 4. **Calculate $P(\text{car or laptop})$:** $$P(\text{car or laptop}) = 0.78 + 0.47 - 0.31 = 0.94$$ 5. **Calculate $P(\text{neither car nor laptop})$:** Since owning car or laptop and owning neither are complementary events: $$P(\text{neither}) = 1 - P(\text{car or laptop}) = 1 - 0.94 = 0.06$$ **Final answers:** - (A) $0.94$ - (B) $0.06$