Question: From a survey involving 1000 university students, a market research company found that 780 students owned laptops, 400 owned cars, and 370 owned cars and laptops. If a university student is selected at random, what is each (empirical) probability?
(A) The student owns either a car or a laptop
(B) The student owns neither a car nor a laptop
1. **State the problem:**
We have 1000 students surveyed.
- Number owning laptops: $780$
- Number owning cars: $400$
- Number owning both cars and laptops: $370$
We want to find:
(A) Probability student owns either a car or a laptop.
(B) Probability student owns neither a car nor a laptop.
2. **Recall the formula for union of two events:**
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
where $A$ is owning a laptop, $B$ is owning a car.
3. **Calculate probabilities:**
- $P(\text{laptop}) = \frac{780}{1000} = 0.78$
- $P(\text{car}) = \frac{400}{1000} = 0.40$
- $P(\text{both}) = \frac{370}{1000} = 0.37$
4. **Calculate $P(\text{car or laptop})$:**
$$P(\text{car or laptop}) = 0.78 + 0.40 - 0.37 = 1.18 - 0.37 = 0.81$$
5. **Calculate $P(\text{neither car nor laptop})$:**
Since total probability is 1,
$$P(\text{neither}) = 1 - P(\text{car or laptop}) = 1 - 0.81 = 0.19$$
**Final answers:**
- (A) $0.81$
- (B) $0.19$