# Thread: Solving a 8 set Venn diagram

1. ## Solving a 8 set Venn diagram

Hi everyone I'm new and i'm just looking for help in doing a Venn diagram I've been set. I'm doing a bachelor of computer science and I have no idea how this relates to computers I just need some help or some answers. I've tried to attempt the question several times but when I come to check it I get either 99 or 102.

A group of 100 students is polled to see how many watched three TV shows, Action, Buzz
and Calypso. The results showed that 59 watched Action, 68 watched Buzz, 52 watched
Calypso, 41 watched Action and Buzz, 34 watched Action and Calypso, 38 watched Buzz
and Calypso, and 11 did not watch any of the three. Let A denote the set of students who
watched Action, and similarly define sets B and C.

2. ## Re: Solving a 8 set Venn diagram

Originally Posted by jakeypoo
Hi everyone I'm new and i'm just looking for help in doing a Venn diagram I've been set. I'm doing a bachelor of computer science and I have no idea how this relates to computers I just need some help or some answers. I've tried to attempt the question several times but when I come to check it I get either 99 or 102.

This is only three sets, not eight...

3. ## Re: Solving a 8 set Venn diagram

Originally Posted by jakeypoo
Hi everyone I'm new and i'm just looking for help in doing a Venn diagram I've been set. I'm doing a bachelor of computer science and I have no idea how this relates to computers I just need some help or some answers. I've tried to attempt the question several times but when I come to check it I get either 99 or 102.

Hence :

$\displaystyle \begin{cases}59=a+75-x \\68=b+79-x \\52=c+72-x \\89=a+b+c+113-2x\end{cases}$

4. ## Re: Solving a 8 set Venn diagram

Hello, jakeypoo!

A group of 100 students is polled to see how many watched three TV shows: Action, Buzz and Calypso.

The results showed that:
. . 59 watched Action,
. . 68 watched Buzz,
. . 52 watched Calypso,
. . 41 watched Action and Buzz,
. . 34 watched Action and Calypso,
. . 38 watched Buzz and Calypso,
. . and 11 did not watch any of the three.

What is the question?

I assume we are to find the number of students
. . who watched all three shows: $\displaystyle n(A\cap B\cap C)$

We have:
. . $\displaystyle \begin{array}{ccccc} n(U) &=& 100 \\ n(A) &=& 59 \\ n(B) &=& 68 \\ n(C) &=& 52 \\ n(A\cap B) &=& 41 \\ n(A \cap C) &=& 34 \\ n(B \cap C) &=& 38 \\ n(A' \cap B' \cap C') &=& 11 \end{array}$

We know that:

. . $\displaystyle n(A \cup B \cup C) \;=\;n(U) - n(A' \cap B' \cap C') \;=\;100 - 11 \;=\;89$

Formula:

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

. . $\displaystyle 89 \;=\;59 + 68 + 52 - 41 - 34 - 38 + n(A\cap B\cap C)$

. . $\displaystyle 89 \;=\;66 + n(A\cap B\cap C)$

Therefore: .$\displaystyle n(A\cap B\cap C) \;=\;23$

5. ## Re: Solving a 8 set Venn diagram

Originally Posted by princeps

Hence :

$\displaystyle \begin{cases}59=a+75-x \\68=b+79-x \\52=c+72-x \\89=a+b+c+113-2x\end{cases}$
Thanks for the help but I'm not too sure what to do with this, please could you explain further?

Originally Posted by Soroban
Hello, jakeypoo!

I assume we are to find the number of students
. . who watched all three shows: $\displaystyle n(A\cap B\cap C)$

We have:
. . $\displaystyle \begin{array}{ccccc} n(U) &=& 100 \\ n(A) &=& 59 \\ n(B) &=& 68 \\ n(C) &=& 52 \\ n(A\cap B) &=& 41 \\ n(A \cap C) &=& 34 \\ n(B \cap C) &=& 38 \\ n(A' \cap B' \cap C') &=& 11 \end{array}$

We know that:

. . $\displaystyle n(A \cup B \cup C) \;=\;n(U) - n(A' \cap B' \cap C') \;=\;100 - 11 \;=\;89$

Formula:

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

. . $\displaystyle 89 \;=\;59 + 68 + 52 - 41 - 34 - 38 + n(A\cap B\cap C)$

. . $\displaystyle 89 \;=\;66 + n(A\cap B\cap C)$

Therefore: .$\displaystyle n(A\cap B\cap C) \;=\;23$
So with these numbers do I put them In the diagram, I'm not sure what to do. I just have to put the numbers into the diagram. Thank you for the help.

6. ## Re: Solving a 8 set Venn diagram

solve system of equations . find x....

7. ## Re: Solving a 8 set Venn diagram

Originally Posted by jakeypoo
Hi everyone I'm new and i'm just looking for help in doing a Venn diagram I've been set. I'm doing a bachelor of computer science and I have no idea how this relates to computers I just need some help or some answers. I've tried to attempt the question several times but when I come to check it I get either 99 or 102.

You don't really need to solve a system of equations here, it just takes a bit of common sense. Set up your Venn diagram as shown in post 3 (with three overlapping circles).
Now analyse what information you've been given. You know there are 100 students polled.
Of these 11 didn't watch any of the three, so you can put 11 in the space OUTSIDE the circles. This leaves 89 students, which means \displaystyle \displaystyle \begin{align*} n(A \cup B \cup C) = 89 \end{align*}.

You should know that \displaystyle \displaystyle \begin{align*} 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) \end{align*}, and you have enough information to now find \displaystyle \displaystyle \begin{align*} n(A \cap B \cap C) \end{align*}.

\displaystyle \displaystyle \begin{align*} 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) \\ 89 &= 59 + 68 + 52 - 41 - 34 - 38 + n(A \cap B \cap C) \\ 89 &= 66 + n(A \cap B \cap C) \\ 23 &= n(A \cap B \cap C) \end{align*}

So that means that in the intersection of all three sets, you need to put 23.

Now we need to fill in the other intersections. We are told that \displaystyle \displaystyle \begin{align*} n(A \cap B) = 41 \end{align*}, but of those 41, there are already 23 in the intersection of all three sets. So that means that what needs to go in the part of \displaystyle \displaystyle \begin{align*} A \cap B \end{align*} which does NOT include C is 41 - 23 = 18.

We are told that \displaystyle \displaystyle \begin{align*} n(A \cap C) = 34 \end{align*}, but of those 34, there are already 23 in the intersection of all three sets. So that means that what needs to go in the part of \displaystyle \displaystyle \begin{align*} A \cap C \end{align*} which does NOT include B is 34 - 23 = 11.

We are told that \displaystyle \displaystyle \begin{align*} n(B \cap C) = 38 \end{align*}, but of those 38, there are already 23 in the intersection of all three sets. So that means that what needs to go in the part of \displaystyle \displaystyle \begin{align*} B \cap C \end{align*} which does NOT include A is 38 - 23 = 15.

Finally we need to fill in the remainders of A, B and C.

We are told that \displaystyle \displaystyle \begin{align*} n(A) = 59 \end{align*}, but of those 59, there are 18 + 11 + 23 = 52 which are already in the intersections which include A. That means that there are 59 - 52 = 7 that are ONLY in A.

We are told that \displaystyle \displaystyle \begin{align*} n(B) = 68 \end{align*}, but of those 68, there are 18 + 15 + 23 = 56 which are already in the intersections which include B. That means that there are 68 - 56 = 12 that are ONLY in B.

We are told that \displaystyle \displaystyle \begin{align*} n(C) = 52 \end{align*}, but of those 52, there are 11 + 15 + 23 = 49 which are already in the intersections which include C. That means that there are 52 - 49 = 3 that are ONLY in C.

8. ## Re: Solving a 8 set Venn diagram

Thank you very much