Discrete mathematics - venn diagram logic

Question

An anonymous survey of college students that determined their behaviors regarding alcohol, cigs, and illegal drugs. Results:

• 894 drank alcohol regularly
• 192 used illegal drugs
• 114 drank alcohol regularly and used illegal drugs
• 97 engaged in all three behaviors
• 665 smoked cigs
• 424 drank alcohol regularly and smoked cigs
• 119 smoked cigs and used illegal drugs
• 309 engaged in none of the behaviors

a) find the number of students in the survey

b) find the number of students who engaged in exactly two of these these behaviors

I drew a venn diagram for this question.

• part a) I used all the students used in this diagram, it is incorrect.
• part b) I subtracted what I did for a by the number of students who did two or more.

both of these parts are incorrect I believe.

Answer 1

Instead of 114 write 114-97=17 similarly for all others except 97 that is correct.

See the question says that 114 drank alcohol regularly and that will include the intersection of three circles also.

Answer 2

Your Venn diagram is divided into eight segments:

Let $D$ be the segment of people that only do drugs. They do not do alcohol or cigarettes. You've written that as $192$. That is wrong. $192$ is everybody who does drugs whether or not they do alcohol or cigarettes. Six of your eight segments are wrong.

Let $AD$ be the segment of people that do alcohol and drugs but not cigarettes.

Let $A$ be the segment of people that do only alcohol.

Let $DC$ be the segment of people that do drugs and alcohol only.

Let $ADC$ be the segment of people that do all three. This is the only internal segment you did right.

Let $AC$ be the segment of people that do Alochol and cigarettes only.

Let $C$ be the segment of people that did only cigarettes.

Let $X$ be the segment of the people that did none. You did that one right.

"894 drank alcohol regularly". That means $A + AD + AC + ADC= 894$

"192 used illegal drugs" That means $D + AD + DC + ADC = 192$

"114 drank alcohol regularly and used illegal drugs" That means $AD + ADC = 114$.

"97 engaged in all three behaviors" That means $ADC = 97$

"665 smoked cigs" That means $C + AC + DC + ADC = 665$

"424 drank alcohol regularly and smoked cigs" That means $AC + ADC = 424$

"119 smoked cigs and used illegal drugs" That means $DC + ADC = 119$

"309 engaged in none of the behaviors" That means $X = 309$.

Now redo the entire thing.

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894 drank alcohol regularly
192 used illegal drugs
114 drank alcohol regularly and used illegal drugs
97 engaged in all three behaviors
665 smoked cigs
424 drank alcohol regularly and smoked cigs
119 smoked cigs and used illegal drugs
309 engaged in none of the behaviors

Your mistake was in calculationg $|A \cap D \cap C^c| = 114$. You assumed incorrectly that $A \cap D$ (those who do Alcohol and Drugs) and $A \cap D \cap C^c$ (those who do Alcohol and Drugs only; no cigarettes) where the same. They are not.

$A \cap D \cap C^c \subset A \cap D$ and $A \cap D = (A \cap D \cap C) \cup (A \cap D \cap C^c)$

So $|A \cap D| = |A \cap D \cap C| + |A \cap D \cap C^c| - |(A \cap D \cap C) \cap (A \cap D \cap C^c)|$

$114 = 97 + |A \cap D \cap C^c| + |\emptyset|$

$|A \cap D \cap C^c| = 17$.

That was your entire problem. But you didn't have to break the entire venn diagram down, if you use the inclusion/exclusion principle.

($|A \cup B| = |A| + |B| - |A \cap B|$ which can inductively extended to $|A \cup B \cup C| = |A|+|B| +|C| - |A \cap B|-|A \cap C|-|B \cap C|+|A \cap B\cap C|$)

$|A| = 894$

$|D| = 192$

$|A \cap D| = 114$

[This means $|A \cup D| = |A| + |D| - |A\cap D| =894 + 192 - 114 = 927$ just in case that is ever asked...]

$|A \cap D \cap C| = 97$

$|C| = 665$

$|A \cap C| = 424$

$|C \cap D| = 119$

$|(A \cup C \cup D)^c| = 309$

All students = $|A \cup C \cup D| + |(A \cup C \cup D)^c|$

$|(A \cup C \cup D)^c| = 309$

$|A \cup C \cup D| = |A| + |C| + |D| - |A \cap C| - |A \cap D| - |C \cap D| + |A \cap C \cap D|$

$|A| = 894; |C|=665; |D|=192; |A \cap C|=424; |A \cap D| = 114; |C \cap D|=119;|A \cap C \cap D| = 97$

So students are $309 + 894 + 665 + 192 - 424 - 114 - 119 + 97$.

To figure out how many engaged in exactly two you do:

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People who engage in exactly two = people engaged in at least to minus those engage in all three.

Those in A and D only are $|A \cap D| - |A \cap D \cap C|=114-97 = 17$

Those in A and C only are $|A \cap C| - |A \cap D \cap C| =424 - 97 = 327$

Those in D and C only are $|D \cap C|- |A\cap D\cap C| = 119 - 97 = 22$

So those in exactly two are $17+327 + 22$.