Question

From a survey involving​ 1,000 university​ students, a market research company found that 710 students owned​ laptops, 470 owned​ cars, and 320 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.

Answer 1

The probability that a student owns either a car or a laptop is 0.86. The probability that a student owns neither a car nor a laptop is 0.14.

Step-by-step explanation:

(A) The student owns either a car or a laptop:

To find the probability that a student owns either a car or a laptop, we need to find the number of students who own a car, the number of students who own a laptop, and the number of students who own both. We can use the formula:

P(A or B) = P(A) + P(B) - P(A and B)

Given that 470 students own cars, 710 students own laptops, and 320 students own both, we can substitute the values into the formula:

P(car or laptop) = P(car) + P(laptop) - P(car and laptop)

P(car or laptop) = 470/1000 + 710/1000 - 320/1000

P(car or laptop) = 860/1000

P(car or laptop) = 0.86

Therefore, the probability that a student owns either a car or a laptop is 0.86.

(B) The student owns neither a car nor a laptop:

To find the probability that a student owns neither a car nor a laptop, we need to subtract the probability that a student owns either a car or a laptop from 1:

P(neither car nor laptop) = 1 - P(car or laptop)

P(neither car nor laptop) = 1 - 0.86

P(neither car nor laptop) = 0.14

Therefore, the probability that a student owns neither a car nor a laptop is 0.14.

Answer 2

Answer: ​(A) 0.86

(B) 0.14

Step-by-step explanation:

Given : Total university students n(S)= 1,000

Number of students owned​ laptops n(L)= 710

Number of students owned cars n(C)= 470

​Number of students owned​ laptops and cars n(L∩C)= 320

Then, the number of students owned either a car or a laptop ​:

n(L∪C)=n(L)+n(C)-n(L∩C)

=710+470-320=860

Probability that the student owns either a car or a laptop ​ :-

`P(L\cup C)=(n(L\cup C))/(n(S))\\\\=(860)/(1000)=0.86`

Probability that the student owns neither a car nor a laptop :-

`P(L'\cap C')=P(L\cup C)'\ [\because (A'\cap B)'=(A\cap B)']\\\\=1-P(L\cup C)\ \ \text{(Using Demorgon's law)}\\\\=1-0.86=0.14`