SOLUTION: 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 an

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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?
Found 2 solutions by jim_thompson5910, ikleyn:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Problem

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
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?


Draw out a Venn Diagram which consists of a rectangle and inside the rectangle will go 3 overlapping circles as shown below

S = set of people who own smartphones
T = set of people who own tablets
C = set of people who own computers
U = universal set = set of all people surveyed

There are 8 distinct regions of this Venn Diagram. The regions are (a) through (h) and are defined as such:



Let's list out the facts that we are given

We won't be using fact 1 to solve this current problem. If you're curious about it, then this represents the sum of regions (h), (g), (f), and (c) added up. In other words, it is the total sum of people outside of circle S.


From fact 2, we can fill in region (a) with 18%
From fact 3, we can fill in region (b) with 29%
From fact 4, we can fill in region (d) with 21%
From fact 5, we can fill in region (e) with 15%



We're being asked "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?".

This is a two part question
Part A: what is the probability that the person would own a smartphone only
Part B: what is the probability that the person would own a smartphone and computer only?

The answer to part A is 18% as this is region (a) found earlier. The answer to part B is 21% which is the value in region (d).
Adding these percentages up gets us 18+21 = 39%. We can add the percentages like this because the regions (a) through (h) are mutually exclusive regions.



So 39% of the people surveyed either own a smartphone only, or they own a smartphone and a computer only.

The probability of randomly selecting a person that either owns a smartphone only, or they own a smartphone and a computer only is 39%. In decimal form that would convert to 0.39, and as a fraction it would be 39/100. These all represent the same answer, just written in a different form.

Answer by ikleyn(52779)   (Show Source): You can put this solution on YOUR website!
.

            Honestly, I don't know and don't understand  WHY  the respectful tutor Jim chose this complicated way to solve the problem.

            It can be solved in  MUCH  SIMPLER  way,  and I will show it to you now.


You have the universal set of all people surveyed.

        Notice that  17% + 18% + 29% + 21% + 15% = 100%,  so these subsets cover the entire set.

Now,  from the text,  it should be clear to you that all listed categories of people are  DISJOINT :
the intersections between any two different categories are  EMPTY.

        It is clear and obvious from the definitions of these categories in the post.


Now,  the question is :   what is the probability to randomly select from the union of the  {18%}  and  {21%}  subsets.

But of course,  this probability is the sum  18% + 21% = 39%.          ANSWER


It is a  DIRECT  CONSEQUENCE  that the given categories 

    a)  cover the entire universal set,     and that

    b)  the categories are disjoint, i.e. have empty intersections.

It is fully consistent with the general formula of the Elementary probability theory
        P(A U B) = P(A) + P(B)
for the disjoint events.

My solution is  completed  at this point.


----------------

A good style educational / (teaching)  tradition assumes and requires that used teaching tools should not
be more complicated than the problem itself.

Or,  in other words,  the solution should be  AS  SIMPLE  AS  POSSIBLE   //   still remaining to be correct.


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