Carcass wrote:
Among a population of 24,000 people, 80 percent own a personal computer, 88 percent own a car, and 72 percent own both a personal computer and a car. If 1 person is to be randomly selected from the 24,000 people, what is the probability that the person selected will be one who owns a car but does NOT own a personal computer?
A. 22/25
B. 1/2
C. 4/25
D. 3/25
E. 2/25
(Use a venn diagram)
Total = PC + Car - Both PC and Car + None
\(24000 = 19200 + 21120 - 17280\) + None
None = 960
Person who owns a Car but not PC = Car - Both PC and Car = \(21120 - 17280 = 3840\)
Probability = \(\frac{3840}{24000} = 0.16 = \frac{4}{25}\)
Hence, option C