# Sets and counting

1. Jul 29, 2014

### jonroberts74

1. The problem statement, all variables and given/known data

A study was done to determine the efficacy of three different drugs – A, B, and C – in relieving headache pain. Over the period covered by the study, 50 subjects were given the chance to use all three drugs. The following results were obtained:

21 reported relief from drug A
21 reported relief from drug B
31 reported relief from drug C
9 reported relief from both drugs A and B
14 reported relief from both drugs A and C
15 reported relief from both drugs B and C
41 reported relief from at least one of the drugs.

b. How many people got relief from none of the drugs?
c. How many people got relief from all three drugs?
d. How many people got relief from A only?
e. Fill in the full Venn diagram for this data.

3. The attempt at a solution

I drew a venn diagram,

running into an issue though.

21 people reported relief from drug A, 9 reported relief from A and B and 14 report relief from A and C. so $A \cap B = 9; A \cap C = 14$ but A only has 21, 21 -14-9 = -2 thats not yet counting $A \cap B \cap C$

similarly for B, B has 21 reports of relief and $B \cap A =9; B \cap C =15$ so 21 - 15 -9 = -3

C seems to have no issue so far, it'll have 2 people that potentially only had relief from C

Last edited: Jul 29, 2014
2. Jul 29, 2014

### LCKurtz

But some of these are the same people; you are counting them twice.

Also, perhaps you should use $n(A\cap B)$ for the number of people in that set. Look in your text for a formula for $n(A\cup B\cup C)$. That will help you figure out $n(A\cap B \cap C)$.

3. Jul 29, 2014

### jonroberts74

$n(A\cup B\cup C) = n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap C) - n(B\cap C) + n(A\cap B \cap C)$

$n(A) + n(B) + n(C) = 21 + 21 + 31$

$- n(A\cap B) - n(A\cap C) - n(B\cap C) = -9 - 14 -15$

$n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap C) - n(B\cap C) = 21 + 21 + 31 - 9 - 14 -15 = 35$

then it says at least 41 report relief from at least one drug

so $n(A\cap B \cap C) = 41 - \Bigg[n(A) + n(B) + n(C) - n(A\cap B) - n(A\cap C) - n(B\cap C)\Bigg] = 41-35 = 6$

D) A only has 4, B only has 3, C only has 8, $n(A \cap B \cap C)=6, n(A \cap B) = 3, n(A \cap C) = 8, n(B \cap C) = 9$, 9 left outside the three