Question

a survey of 500 television watchers

produced the information:
285 watch football games
195 watch hockey games
115 watch basketball games
45 watch football and basketball
70 watch football and hockey
50 watch hockey and basketball
50 do not watch any of the three games

A. how many people in the survey watch all the three games?

B. How many people watch exactly one of the three games?

Answer 1

In venn diagram

Answer 2

The principle of inclusion-exclusion would take care of this...but im a bit fuzzy on the finer details....one sec, going through old notes lol

Answer 3

i think A is 310

Answer 4

we have 500 people
the sum of number of elements in each set total 810

Answer 5

so we have some overlapping

Answer 6

does anyone think anything of my answer to A

Answer 7

maybe i should have done 810-450

Answer 8

im checking it right now, actually, let me upload the rules for the inclusion exclusion principle with some examples, you could probably understand it a little better than i can lol

Answer 9

i will look it up

Answer 10

im getting 20 for the answer lol

Answer 11

for the number of people that watch all 3

Answer 12

THE NUMBER OF PEOPLE WHO WATCH ALL 3 IS 10.

Answer 13

its:
500-285-195-115+70+45 +50 - 50

Answer 14

im 100% sure its 20 now

Answer 15

Heres how Inclusion Exclusion works:
Let N(a) be the number of people that watch football, N(b) for hockey, N(c) for basketball.
N(ab) for football and hockey, N(ac) for football and basketball, N(bc) for hockey and basketball, N(abc) for all three (what we want), and N(a'b'c') for those that watch none of those three.

Answer 16

The formula is:
N(a'b'c') = (Total Number of people) -N(a)-N(b)-N(c)+N(ab)+N(ac)+N(bc)-N(abc)

Answer 17

football watchers=F
Hockey watchers=H
Basketball watchers=B
|F n H n B|=|F|+|H|+|B|-|F U H U B| -|F n B|- |F n H|-|H n B|
=285+195+115-|F U H U B| -70-45-50
=430-|F U H U B|
what is |F U H U B|=?

Answer 18

so we end up with:
50 = 500-285-195-115+70+45+50- N(abc)

Answer 19

solving for N(abc) gives 20

Answer 20

285+195+115-450=145 WHO ARE THE COMMON WATCHERS OF DIFFERENT GAMES.
50-X+X+70-X+45-X=145
X=10

Answer 21

This is the rules of the principle, uploading some examples in a sec.

Answer 22

er, an example. im still 100% sure the answer is 20, its a straightfoward application of this.

Answer 23

here i think found an example havent read through it but i think it will help
http://www.math.uic.edu/~fye/Math118Fall2010/Practice-Final-with-Answers.pdf

Answer 24

no i dont like that example that i posted

Answer 25

i can answer c of that example, but not the other 2, at least in a straightforward manner anyways

Answer 26

ok i think i agree with joe
he showed proof for his formula
and i believe i get it

Answer 27

Thank you :) after that spherical triangle mess i was thinking i wasnt going to get through to anyone anymore ;;
i still dont know how to answer b of this question yet though lol >.<

Answer 28

so what is the not exactly one

Answer 29

at least one right?

Answer 30

i think im thinking of probability

Answer 31

remember this
P(C)=1-P(not C)

Answer 32

right right

Answer 33

ok but i know we arent talking about probability
but lets think
C1 and C2 are events of C
P(C)=1
P(C1)=P(C)-P(not C1)
so does this hold
C1=C-notC1

Answer 34

im gonna stare at that example you posted earlier, i think that would help

Answer 35

for exactly two in there example they got 2+11+1
let me think where did they get the numbers

Answer 36

i think i see it

Answer 37

(285-20)+(195-20)+(115-20)

Answer 38

20 comes from our answer from a right?

Answer 39

nope thats not right lol

Answer 40

well they
have
(7-5)+(16-5)+(6-5) right? let me look back

Answer 41

yes joe 20 came from our answer above

Answer 42

the info for all number who enjoyed all 3 was given to us in the example
and 7,16,6 were the people who enjoyed both of something

Answer 43

the number that enjoyed all was 5

Answer 44

so 7-5, 16-5, and 6-5 gave the number of people that enjoyed only 2 activities

Answer 45

right!

Answer 46

i was think if we took it away from the groups who enjoyed 1 thing we would have it

Answer 47

so 2, 11, and 1, or 14 people enjoyed only 2, 5 people enjoyed 3, so how can we use this to figure out how many people only enjoyed one?

Answer 48

http://www.math.fsu.edu/~wooland/hm2ed/Part1Module3/Part1Module3.pdf

Answer 49

19 people didnt enjoy anything (used the formula above to get that)

Answer 50

ah i think i got it

Answer 51

83 people surveyed. Lets take out the number of people that liked nothing, exactly 2 things, or 3 things. that should leave the number of people that only like exactly one thing.
83 - 19 - 14 - 5 = 45
which is the sum of the numbers in that venn diagram of people that only liked one sport, 15+17+13

Answer 52

So now to apply it to this problem....

Answer 53

it would be the 500 people total, minus the people that watch nothing, the people that watch everything, and the people that watch only 2 things

Answer 54

we gotta calculate the number of people that only watch 2 things, but we figured that out already earlier, its:
70-20+45-20+50-20 = 105

Answer 55

so the number of people that only watch one thing is:
500-50-20-105 = 325

Answer 56

So.....after all this deliberation, im sticking with a) 20 , b) 325

Answer 57

how can i get 20??

Answer 58

hey you might like this site prince
http://interactives.mped.org/view_interactive.aspx?id=28&title=

Answer 59

>.> lol
let x be the number of people that watch all 3, then you need to solve this equationL
50 = 500 - 285 - 195-115+70+45+50-x

Answer 60

ooo that is cool lol

Answer 61

gj joe
im still not sure about b

Answer 62

but joe is really smart so he is probably right

Answer 63

he has probability of 91/100 of being right

Answer 64

oh what i just got 325

Answer 65

i drew a venn diagram and played it with
and i did 190+40+95

Answer 66

nice nice, does this confirm the answers then?

Answer 67

i have 190 only watch football
i have 40 only watch basketball
i have 95 only watch hockey
i have 20 only watch all 3
i have 50 only watch both Hockey and football
i have 30 only watch both basketball and and hockey
i have 25 only watch both fooball and basketball

Answer 68

now we could answer any question lol, nice job :)

Answer 69

yep yep

Answer 70

awesome job joe

Answer 71

another application of inclusion exclusion: how many numbers from 1 to 100 arent divisible by 2, 3, or 5?

Answer 72

this was one of your projects?

Answer 73

i went to a seminar on it and got that packet i posted (had some other examples and problems). i dont know it there is a class that uses this, maybe a combinatorics class? since it involves counting things in an efficient manner?

Answer 74

This deserves breakfast, im gonna go get breakfast <.< i suggest you all do the same lol.