document.write( "Question 1151297: A smartphone company found in a survey that 17% of people did not own a smartphone, 18% owned a smartphone only, 29% owned a smartphone and only a tablet, 21% owned a smartphone and only a computer, and 15% owned all three. If a person were selected at random, what is the probability that the person would own a smartphone only or a smartphone and computer only? \n" ); document.write( "
Algebra.Com's Answer #773026 by ikleyn(52780)![]() ![]() You can put this solution on YOUR website! .\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " Honestly, I don't know and don't understand WHY the respectful tutor Jim chose this complicated way to solve the problem.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " It can be solved in MUCH SIMPLER way, and I will show it to you now.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "You have the universal set of all people surveyed.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " Notice that 17% + 18% + 29% + 21% + 15% = 100%, so these subsets cover the entire set.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now, from the text, it should be clear to you that all listed categories of people are DISJOINT : \n" ); document.write( "the intersections between any two different categories are EMPTY.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " It is clear and obvious from the definitions of these categories in the post.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now, the question is : what is the probability to randomly select from the union of the {18%} and {21%} subsets.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "But of course, this probability is the sum 18% + 21% = 39%. ANSWER\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "It is a DIRECT CONSEQUENCE that the given categories \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( " a) cover the entire universal set, and that\r\n" ); document.write( "\r\n" ); document.write( " b) the categories are disjoint, i.e. have empty intersections.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "It is fully consistent with the general formula of the Elementary probability theory\r \n" ); document.write( "\n" ); document.write( " P(A U B) = P(A) + P(B) \r \n" ); document.write( "\n" ); document.write( "for the disjoint events.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "My solution is completed at this point.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "----------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A good style educational / (teaching) tradition assumes and requires that used teaching tools should not \n" ); document.write( "be more complicated than the problem itself.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Or, in other words, the solution should be AS SIMPLE AS POSSIBLE // still remaining to be correct.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |