SOLUTION: From a survey involving 1,000 university students, a market research company found that 750 students owned laptops, 410 owned cars, and 370 owned cars and laptops.
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-> SOLUTION: From a survey involving 1,000 university students, a market research company found that 750 students owned laptops, 410 owned cars, and 370 owned cars and laptops.
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Question 1197700: From a survey involving 1,000 university students, a market research company found that 750 students owned laptops, 410 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 Answer by ikleyn(52786) (Show Source):
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From a survey involving 1,000 university students, a market research company found
that 750 students owned laptops, 410 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
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(A) This question asks "either a car or a laptop, but not both".
We compute this amount as
(n(car) + n(laptop) - n(both)) - n(both) = (750+410-370) - 370 = 420.
Thus the probability is P(question A) = = = = 0.42 = 42%. ANSWER
(B) This question asks about the complementary probability to having a car OR a laptop.
The number of students who own a car or a laptop is
n(car) + n(laptop) - n(both) = 750 + 410 - 370 = 790.
Thus the probability is P(question B) = = = 0.21 = 21%. ANSWER
Solved.
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In this problem, the key subject to learn is the difference in calculating " either - or " from calculating single " or ".
