Question 1194404: A survey of 130 pet owners yielded the following information: 47 own fish; 53 own a bird; 50 own a cat; 64 own a dog; 2 own all four; 11 own only fish; 14 own only a bird; 10 own fish and a bird; 21 own fish and a cat; 24 own a bird and a dog; 27 own a cat and a dog; 3 own fish, a bird, a cat, and no dog; 1 owns fish, a bird, a dog, and no cat; 9 own fish, a cat, a dog, and no bird; and 10 own a bird, a cat, a dog, and no fish. How many of the surveyed pet owners have no fish, no birds, no cats, and no dogs? (They own other types of pets.)
Found 3 solutions by math_tutor2020, ikleyn, MathTherapy:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
A = person owns a fish
B = person owns a bird
C = person owns a cat
D = person owns a dog
There are 4 pets, so there are 2^4 = 16 different possibilities.
Here are the 16 possibilities- Person doesn't own any of the four pets mentioned
- A - persons owns a fish only
- B - they own a bird only
- C - owns a cat only
- D - owns a dog only
- AB - fish and bird only
- AC - fish and cat only
- AD - fish and dog only
- BC - bird and cat only
- BD - bird and dog only
- CD - cat and dog only
- BCD - bird, cat, and dog (i.e. everything but a fish)
- ACD - fish, cat and dog (i.e. everything but a bird)
- ABD - fish, bird and dog (everything but a cat)
- ABC - fish, bird and cat (everything but a dog)
- ABCD - person owns all four pets mentioned
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Let's draw out the Venn Diagram.
Start by drawing two ovals like so. Make sure they overlap.
Create a mirror copy of those two ovals to have a total of four ovals.
Then slide the ovals over until we get this
It may seem like a bit of a mess, so be sure you can break apart the various ovals (or be able to spot the four ovals).
This is why I recommend following these steps to slide the ovals over.
Unfortunately we can't have the ovals separated or else we won't be able to form the overlapped regions needed.
Next, we'll add the 15 labels mentioned in the previous section above. The first label of "none of the 4 pets mentioned" is ignored for now, as it goes outside the four ovals.
Color-coding things helps us separate the various regions (it may be easy to get lost)
- Regions in white are for the people who own exactly one pet only
- Regions in blue are for the people who own exactly two pets only
- Regions in red are for the people who own exactly three pets only
- The region in green is for the people who own all four pets mentioned
Take note how something like region ABC is in ovals A, B and C at the same time. Region ABC is outside oval D.
Similar logic applies to how the other regions are set up.
The region outside the ovals represents people who own none of the four pets mentioned.
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Let's fill out the Venn Diagram with numeric values.
To do so, we'll need to use the given facts:
| Number | Statement | | 1 | 47 own fish | | 2 | 53 own a bird | | 3 | 50 own a cat | | 4 | 64 own a dog | | 5 | 2 own all four | | 6 | 11 own only fish | | 7 | 14 own only a bird | | 8 | 10 own fish and a bird | | 9 | 21 own fish and a cat | | 10 | 24 own a bird and a dog | | 11 | 27 own a cat and a dog | | 12 | 3 own fish, a bird, a cat, and no dog | | 13 | 1 owns fish, a bird, a dog, and no cat | | 14 | 9 own fish, a cat, a dog, and no bird | | 15 | 10 own a bird, a cat, a dog, and no fish |
We start with statement 5, which says "2 own all four"
This number goes inside the region labeled ABCD.
In other words, this number goes in the region that is inside all four ovals.
Now move onto statement 6. It says "11 own only fish", which means 11 goes in the region labeled "A".
This is the region inside oval A, but outside any other oval.
Now onto statement 7. It says "14 own only a bird", so we'll write "14" in region B. This is the region inside oval B, but outside any other oval.
Let's jump ahead to statement 12.
It says "3 own fish, a bird, a cat, and no dog".
We'll write the number 3 inside the region labeled "ABC".
This is the region inside ovals A,B,C but outside oval D.
To be more accurate, I'll write it off to the side since region ABC is very small.
I'll be doing the same with regions ABD, AC and BD as well.
The next statement (statement 13) says "1 owns fish, a bird, a dog, and no cat"
The number "1" goes with region ABD.
Up next is the statement "9 own fish, a cat, a dog, and no bird" (statement 14). It tells us that the number 9 goes inside region ACD.
Up next is the statement "10 own a bird, a cat, a dog, and no fish" (statement 15).
It tells us that the number 10 goes inside region BCD.
Here's a mini recap so far:
A = 11
B = 14
ABC = 3
ABD = 1
ACD = 9
BCD = 10
ABCD = 2
That takes care of 7 regions, so there are 16-7 = 9 more to go.
This is what your Venn Diagram should look like at this point (after filling in the proper regions)
We'll move to statement 8 now.
It says "10 own fish and a bird"
Use a pen tool to highlight oval A and oval B as shown below.
The two ovals highlighted overlap in this teardrop shaped region marked below
Everything in this teardrop region is in oval A and oval B at the same time. The subregions involved in this teardrop are:
Region AB
Region ABC
Region ABD
Region ABCD
Each of the four strings starts with "AB", and we either leave it as is, or we tack on various combos of C and/or D.
This is a methodical way to list out all the possibilities of someone owning a fish and a bird (and possibly a cat and/or dog).
Algebraically we can form the equation
AB+ABC+ABD+ABCD = 10
since 10 people own a fish and a bird (and possibly a cat and/or dog).
Be careful and try not to think about something like "ABC" as "A*B*C".
ABC is one variable name, and not the product of three separate variables.
We'll apply the following substitutions
ABC = 3
ABD = 1
ABCD = 2
to go from this
AB+ABC+ABD+ABCD = 10
to this
AB+3+1+2 = 10
and it solves to
AB = 4
This says "4 people have a fish and a bird, but not a cat and not a dog".
That was all for statement 8.
Now onto statement 9.
We'll have this equation
AC+ABC+ACD+ABCD = 21
Each item on the left hand side has "A" and "C" in it somewhere.
Plug in these items
ABC = 3
ACD = 9
ABCD = 2
to go from this
AC+ABC+ACD+ABCD = 21
to this
AC+3+9+2 = 21
which solves to
AC = 7
Therefore, 7 people have a fish and a cat, but none of the other two mentioned pets.
Onto statement 10.
We'll have this equation
BD+ABD+BCD+ABCD = 24
Each item on the left hand side has "B" and "D" in it somewhere.
Plug in these items
ABD = 1
BCD = 10
ABCD = 2
to go from this
BD+ABD+BCD+ABCD = 24
to this
BD+1+10+2 = 24
which solves to
BD = 11
Therefore, 11 people have a bird and a dog, but none of the other two mentioned pets.
Onto statement 11.
We'll have this equation
CD+ACD+BCD+ABCD = 27
Each item on the left hand side has "CD" in it.
Plug in these items
ACD = 9
BCD = 10
ABCD = 2
to go from this
CD+ACD+BCD+ABCD = 27
to this
CD+9+10+2 = 27
which solves to
CD = 6
Therefore, 6 people have a cat and a dog, but none of the other two mentioned pets.
Let's update that previous recap:
ABCD = 2
A = 11
B = 14
ABC = 3
ABD = 1
ACD = 9
BCD = 10
AB = 4
AC = 7
BD = 11
CD = 6
The new items are in red.
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Go back to the top to statement 1.
47 people own a fish. Some of these 47 people own a fish only, while others may own exactly one other pet along with the fish, or two other pets along with the fish, or own all 4 pets mentioned.
A = fish owners only
AB = fish + bird
AC = fish + cat
AD = fish + dog
ABC = fish + bird + cat
ABD = fish + bird + dog
ACD = fish + cat + dog
ABCD = all four pets mentioned
It may help to use your pen tool to circle all of oval A.
We have the following values:
A = 11 (given from statement 6)
AB = 4 (solved earlier; see statement 8)
AC = 7 (solved earlier; see statement 9)
AD = unknown, we'll be solving for it soon
ABC = 3 (solved earlier; see statement 12)
ABD = 1 (solved earlier; see statement 13)
ACD = 9 (solved earlier; see statement 14)
ABCD = 2 (given; see statement 5)
These values must add up to the 47 fish owners
A+AB+AC+AD+ABC+ABD+ACD+ABCD = 47
11+4+7+AD+3+1+9+2 = 47
AD+37 = 47
AD = 47-37
AD = 10
There are 10 people who have a fish and a dog, but no other pet mentioned.
Move onto statement 2, which says "53 own a bird".
We'll circle the regions that involve the letter B in some fashion, i.e. we'll circle all of oval B.
B = bird owners only
AB = fish + bird
BC = bird + cat
BD = bird + dog
ABC = fish + bird + cat
ABD = fish + bird + dog
BCD = bird + cat + dog
ABCD = all four pets mentioned
We'll be using these values to help solve for variable BC
B = 14 (given; see statement 7)
AB = 4 (solved earlier; see statement 8)
BD = 11 (solved earlier; see statement 10)
ABC = 3 (solved earlier; see statement 12)
ABD = 1 (solved earlier; see statement 13)
BCD = 10 (solved earlier; see statement 15)
ABCD = 2 (given; see statement 5)
So,
B+AB+BC+BD+ABC+ABD+BCD+ABCD = 53
14+4+BC+11+3+1+10+2 = 53
BC = 8
There are 8 pet owners that have a bird and a cat, but no other pets mentioned.
Now onto statement 3.
Circle oval C, i.e. circle each region that involves C in some way
C = cat only
AC = fish and cat only
BC = bird and cat only
CD = cat and dog only
ABC = fish + bird + cat
ACD = fish + cat + dog
BCD = bird + cat + dog
ABCD = all four pets mentioned
We have the following values:
AC = 7 (solved earlier in statement 9)
BC = 8 (solved earlier in statement 2)
CD = 6 (solved earlier in statement 11)
ABC = 3 (see statement 12)
ACD = 9 (see statement 14)
BCD = 10 (see statement 15)
ABCD = 2 (see statement 5)
So,
C+AC+BC+CD+ABC+ACD+BCD+ABCD = 50
C+7+8+6+3+9+10+2 = 50
C+45 = 50
C = 50-45
C = 5
There are 5 people who own a cat only (and none of the other three pets mentioned).
Now onto the last statement left, which is statement 4.
There are 64 people who own a dog which means,
D+AD+BD+CD+ABD+ACD+BCD+ABCD = 64
D+10+11+6+1+9+10+2 = 64
D+49 = 64
D = 64-49
D = 15
There are 15 dog owners who do not have a fish, nor a bird, nor a cat.
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Here's the complete summary of what we found
A = 11
B = 14
C = 5
D = 15
AB = 4
AC = 7
AD = 10
BC = 8
BD = 11
CD = 6
ABC = 3
ABD = 1
ACD = 9
BCD = 10
ABCD = 2
This is what the completed Venn Diagram looks like
The 14 outside of all of the ovals is the final answer and explained in the next two paragraphs.
Add up all of those values to find out how many people there are that own at least one of the four pets mentioned.
A+B+C+D+AB+AC+AD+BC+BD+CD+ABC+ABD+ACD+BCD+ABCD
11+14+5+15+4+7+10+8+11+6+3+1+9+10+2
116
There are 116 people who have at least one pet of the four pets mentioned.
There are 130 people total.
That must mean there are 130-116 = 14 people who do not have any of the four pets mentioned.
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Final Answer: 14
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
For convenience, I will represent the given part in modified form
by numbering statements for easy referring.
A survey of 130 pet owners yielded the following information:
(1) 47 own fish; 53 own a bird; 50 own a cat; 64 own a dog;
(2) 2 own all four;
(3) 11 own only fish; 14 own only a bird;
(4) 10 own fish and a bird; 21 own fish and a cat; 24 own a bird and a dog; 27 own a cat and a dog;
(5) 3 own fish, a bird, a cat, and no dog;
(6) 1 owns fish, a bird, a dog, and no cat;
(7) 9 own fish, a cat, a dog, and no bird;
(8) 10 own a bird, a cat, a dog, and no fish.
How many of the surveyed pet owners have no fish, no birds, no cats, and no dogs?
(They own other types of pets.)
~~~~~~~~~~~~~~~~~
I present here another solution, which entirely is build on using Inclusion-Exclusion principle.
Simply saying, they want you find the union set of owners, who own either fish, or birds, or cats or dogs,
and then calculate the COMPLEMENT of this set to 130.
It would be a standard Inclusion-Exclusion problem, if the numbers of elements in single sets
F (fish owners), B(bird owners), C (cat owners) and D (dog owners) be given, along with
their in-pairs intersections, triple intersections and quadruple intersection.
But in the problem, the given are other combinations of intersections.
THEREFORE, if you want to reduce the problem to standard Inclusion-Exclusion, you should restore
information about standard intersection subsets, based on given data.
We just have single subsets F, B, C, D given in statement (1), and a quadruple intersetion given in statement (2).
From statements (5), (6), (7), and (8), we can easy restore all triple intersections by adding FBCD = 2
to given values in (5), (6) (7) and (8)
FBC = 3 + 2 = 5; (11)
FBD = 1 + 2 = 3; (12)
FCD = 9 + 2 = 11; (13)
BCD = 10 + 2 = 12. (14)
Regarding in-pair intersections, we just have given FB = 10, FC = 21, BD = 24, CD = 27 in statement (4).
So, of in-pair intersections, we need to restore FD and BC, that are not given in the input.
It is obvious, that the set F - F_only is the union of three subsets FB, FC and FD:
F - F_only = FB U FC U FD; so n(F - F_only) = FB + FC+ FD, or (applying Inclusion-Exclusion)
47 - 11 = FB + FC + FD - FBC - FBD - FCD + FBCD = 10 + 21 + FD - 5 - 3 - 11 + 2.
It gives an equation 36 = 14 + FD, from which FD = 36-14 = 22.
Similarly, the set B - B_only is the union of three subsets FB, BC and BD:
B - B_only = FB U BC U BD; so n(B - B_only) = FB + BC+ BD, or (applying Inclusion-Exclusion)
53 - 14 = FB + BC + BD - FBC - FBD - BCD + FBCD = 10 + BC + 24 - 5 - 3 - 12 + 2.
It gives an equation 39 = 16 + BC, from which BC = 39-16 = 23.
Now the union (F U B U C U D) contains, according to Inclusion-Exclusion principle,
F + B + C + D - FB - FC - FD - BC - BD - CD + FBC + FBD + FCD + BCD - FBCD =
= 47 + 53 + 50 + 64 - 10 - 21 - 22 - 23 - 24 - 27 + 5 + 3 + 11 + 12 - 2 = 116 owners.
The complement of 116 to 130 is 14 owners.
ANSWER. 14 owners have other types of pets.
Solved.
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Notice that in my solution, the meaning of symbols FB, FC, FD, BC, BD, CD, FBC, FBD, FCD, BCD, FBCD
is different from that in the post of the other tutor.
In my post, they mean in-pair, triples and quadruple intersection of the base subsets F, B, C and D.
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On inclusion-exclusion principle, see this Wikipedia article
https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle
To see many other similar (and different) solved problems, see the lessons
- Counting elements in sub-sets of a given finite set
- Advanced problems on counting elements in sub-sets of a given finite set
- Challenging problems on counting elements in subsets of a given finite set
- Selected problems on counting elements in subsets of a given finite set
- Inclusion-Exclusion principle problems
in this site.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! A survey of 130 pet owners yielded the following information: 47 own fish; 53 own a bird; 50 own a cat; 64 own a dog; 2 own all four; 11 own only fish; 14 own only a bird; 10 own fish and a bird; 21 own fish and a cat; 24 own a bird and a dog; 27 own a cat and a dog; 3 own fish, a bird, a cat, and no dog; 1 owns fish, a bird, a dog, and no cat; 9 own fish, a cat, a dog, and no bird; and 10 own a bird, a cat, a dog, and no fish. How many of the surveyed pet owners have no fish, no birds, no cats, and no dogs? (They own other types of pets.)
F = Fish B = Bird C = Cat D = Dog
.png) .
Total # of pet-owners (@ least 1 pet): 4 + 3 + 1 + 2 + 9 + 10 + 11 + 14 + 7 + 11 + 13 + 6 + 25 = 116
Total number of pet-owners with no cats, dogs, fish, or birds: 130 - 116 = 14
BTW: Contrary to what was stated, bird-owners don’t total 53, and fish-owners don’t total 47.
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