[Math] A problem on Venn diagram

elementary-set-theory

Please does the following problem requires three overlapping circles or just two to display the information in a Venn diagram? I need some hints.

A stock broker was presented with the detailed samples of companies that were either in the retail sector or quoted on the Stock Exchange or both. 200 companies were studied (set C). Of these two hundred companies, 140 were retail companies (set R) and 120 were quoted on the stock exchange (set Q). Months later, 90 of these two hundred companies failed (F forms the failed companies), even though only five of these failures were not in the retail sector.

a) Depict the information in a Venn diagram

b) What is the maximum and minimum number of companies studied that failed and were retail companies but not quoted on the Stock Exchange

c) Describe and show on the Venn diagram, the following

i. Quoted retail companies that were not studied

ii. Studied companies that were quoted but did not fail

iii. Studied companies that failed but were neither quoted nor retail companies

Best Answer

You can express the various overlaps with $3$ circles, since there are three states for each of the companies (elements of $C$):

  • Either it is in retail (an element of $R$) or it is not.
  • Either it is quoted in the stock exchange (an element of $Q$) or it is not.
  • Either it failed (is an element of $F$) or it is not.

Added: To start with, we have the following diagram:

enter image description here

We want to find that the maximum and minimum for that unknown value $x$. Since $5$ of the $90$ failed companies were not in retail, then $90-5=85$ were in retail. That is, of the studied companies, there were $85$ that failed and were retail companies. This tells us that $0\le x\le 85$ and that:

enter image description here

The two unfilled regions of $F$ must add up to $5$ (since $5$ of the failed companies were not in retail). Likewise, since $85$ of the $140$ retail companies failed, then $140-85=55$ of them did not, so the two unfilled regions of $R$ must add up to $55$. Hence:

enter image description here

where $0\le y\le 5$ and $0\le z\le 55.$ Now, all told, there are $120$ quoted companies, and we've accounted for $85-x+z+y$ of them so far. Thus:

enter image description here

Now, our total is $180+x-y-z,$ so:

enter image description here

Now, remember that $0\le x\le 85.$ We can refine this just a bit. Of the $200$ studied companies, $85$ of them were retail companies that failed and $120$ of them were companies quoted on the stock exchange. In order to make sure that there is "enough room" for these two sets of companies, we will need at least $120+85-200=5$ companies to be retail companies that failed and were quoted on the stock exchange, meaning $85-x\ge 5,$ meaning $80\ge x.$ Hence, our bounds on $x$ are $0\le x\le 80.$ At this point, we're all set to go. We know that $0\le x\le 80,$ so we have upper and lower bounds. To see whether they are the maximum and minimum, we must see if it is possible for those bounds to be attained (that is, see if it is possible for $x=80$ or $x=0$). The substitution $x=80$ gives us:

enter image description here

Recall that $0\le z\le 55.$ Of the $200$ companies, $140$ are in retail and $120$ were quoted, so to make sure there is "enough room" for those sets of companies, we need at least $140+120-200=60$ retail companies that were quoted. That is, $z+5\ge 60,$ so $z\ge 55.$ Hence, we need $z=55,$ and so:

enter image description here

Similarly, we need at least $90+120-200=10$ companies to be quoted companies that failed, meaning $5+y\ge10,$ so $y\ge 5,$ and since $0\le y\le 5,$ then $y=5,$ so:

enter image description here

Therefore, $80$ is the maximum number attainable for $x$.

Now, let's check and see if $x=0$ can be obtained. Substitution gives us:

enter image description here

Since $y\ge 0$ and $35-y-z\ge 0,$ then $35-z\ge 35-y-z\ge 0,$ so $z\le 35,$ so our new bounds of $z$ (having specified $x$) are $0\le z\le 35.$ Likewise, $y\le 35,$ but this doesn't refine the constraints $0\le y\le 5$ that we already had. Let's just pick a value of $z$ and see what happens. Say $z=10.$ Then:

enter image description here

That didn't introduce any issues, either, nor did it narrow our constraints for $y$ at all, so we can still pick any $0\le y\le 5$ and get an appropriate Venn diagram. Thus, $0$ is the minimum number attainable for $x$.

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