Question 1197700
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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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<pre>
(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) = {{{420/1000}}} = {{{42/100}}} = {{{21/50}}} = 0.42 = 42%.      <U>ANSWER</U>



(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) = {{{(1000-790)/1000}}} = {{{21/100}}} = 0.21 = 21%.      <U>ANSWER</U>
</pre>

Solved.



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In this problem, the key subject to learn is the difference in calculating &nbsp;" either - or&nbsp;" &nbsp;from calculating single &nbsp;" or ".



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;To memorize :


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The single &nbsp;" or " &nbsp;is &nbsp;" or inclusive ".


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The &nbsp;" either-or " &nbsp;is &nbsp;" or exclusive ".