# Probability sports store problem

1. Dec 18, 2006

### dontdisturbmycircles

1. The problem statement, all variables and given/known data
A survey was done to determine whether a sports store should carry stock for skiers. The proportions of people who participate in skiing, skating and racquet sports are as follows.

skiing: 0.40
skating: 0.55
racket sports: 0.35
skiing and skating: 0.05
skiing and racket sports: 0.15
skating and racket sports: 0.15
all 3 sports: 0.05

What is the probability that a person chosen at random only participates at skiing

2. Relevant equations

3. The attempt at a solution
I want to say that it is

P(only skiing) = P(skiing) - P(skiing and skating) - P(skiing and racket sports)

, that would be all fine and dandy if people only played 2 sports, but then I have the all 3 option... does the "skiing and racquet sports" include people who play all three? I would find it hard to believe since all people whom ski and skate would NEED to also play racket sports in order for the all 3 sports to be true. But can that be ruled out?

I am basically just thinking as I post so if it doesn't make sense, excuse it please. :P

Argggggh!!! The probability of me going nuts is nearing 1 at this point. lol

Last edited: Dec 18, 2006
2. Dec 18, 2006

### dontdisturbmycircles

Nevermind I think I have it...

P(skiing only) = p(skiing)-p(skiing and skating) - p(skiing and racket) - p (all 3) since the ones like "skiing and skating" must mean skiing, skating and not racket sports..

I dunno what is with this probability stuff, I just am not that great at it... I'll have to practice more.

Last edited: Dec 18, 2006
3. Dec 18, 2006

### quasar987

First let's label the events with letters for brievety. For instance call

E1: a given person likes skiing
E2: a given person likes skating
...

In set language then, what event are you looking for? You want the probability that

A: a given person likes skiing but does not like skiing and skating nor skiing and rackets nor skiing and raquet and skating

So

$$A=E_1\cap E_4^c\cap E_5^c\cap E_7^c$$

Then you gotta find how to find the probability of that with what you know. Hint: Poincaré's formula.

4. Dec 18, 2006

### dontdisturbmycircles

I see what you are getting at. Although this is a different way of thinking about it then taught in my highschool course, I am still interested since it makes sense. All of these probabilities make up sets. Thus I am interested in the probability that a value lies in set 1 but not set 4 5 or 7.

I would like to avoid using Poincare's formula if possible since I haven't learned it yet. Although I am interested. How exactly would one go about finding out the probability that a value lies in set 1 but not set 4 5 or 7... I am still thinking about it.

5. Dec 18, 2006

### dontdisturbmycircles

I think I am right... if 50% of people like the letter B and 25 % like only B and C and 7% like B and C and D then 0.5-0.25-0.07=0.18 is the probability that a randomly selected person only likes B

Since if I take 100 people, and put them in barrels (sets?) 50 people will be placed in barrel B but then 25 people will say they belong in barrel "B and C" and 7 people will say they belong in barrel "B C and D".... so I have 18 people left in barrel B.

Lol, okay yea, I am right. Sorry for using this thread as a think tank. lol

Last edited: Dec 18, 2006
6. Dec 18, 2006

### quasar987

Be careful. When they write that the proportion of people who like skiing and skating is 0.05, it does not mean that the proportion of people only like skiing and skating. Some of them might enjoy racket sport as well.

In the set notation I introduced in post #3,

(people who like skiing and skating)=$E_1\cap E_2$

Do you mean to say you don't use a set approach at all in your class?!

7. Dec 18, 2006

### quasar987

Because you said you were interested, in set notation, the solution is as follow:

$$A=E_1\cap E_4^c\cap E_5^c\cap E_7^c=(E_1^c\cup E_4\cup E_5\cup E_7)^c$$

(By DeMorgan's laws)

$$\Rightarrow P(A)=P((E_1^c\cup E_4\cup E_5\cup E_7)^c)=1-P(E_1^c\cup E_4\cup E_5\cup E_7)$$
$$=P(E_1^c)+P(E_4)+P(E_5)+P(E_7)-P(E_1^c\cap E_4)-P(E_1^c\cap E_5)-P(E_1^c\cap E_7)-P(E_4\cap E_5)-P(E_4\cap E_7)-P(E_5\cap E_7)$$
$$+P(E_1^c\cap E_4\cap E_5)+P(E_1^c\cap E_4\cap E_7)+P(E_1^c\cap E_5\cap E_7)+P(E_4\cap E_5\cap E_7)-P(E_1^c\cap E_4\cap E_5\cap E_7)$$

etc.

Most of the terms are 0 because for instance, $E_1^c\cap E_4$ is the set of all people who don't like skiing but who like skiing and skating. This is empty cuz either you like skiiing or you dont.

Last edited: Dec 18, 2006
8. Dec 18, 2006

### dontdisturbmycircles

pretty much. we started by reviewing the counting rule. Then moved onto permutations and combinations and factorial notation. Then when we are solving probability questions we started using sample spaces but then moved onto using the addition rule (for probability questions with the word OR) and the multiplication rules (for the word and)... They actually called one of the multiplication rules (P(A and B) = P(A)*P(B|A) "Bayes' formula"... I don't know whether or not this is an appropriate name for it but I had to learn Bayes' Theorem $$P(A|B)=\frac{P(A)*P(B|A)}{P(A)*P(B|A)+P(\overline{A})*P(B|\overline{A})}$$

on my own.. I picked up a probability textbook from the library but haven't really had time to read a lot of it. Have been reading a university level calculus book as well as browsing a physics with calculus university level text... and doing that alone on top of normal HS stuff is hard enough.

Last edited: Dec 18, 2006
9. Dec 18, 2006

### drpizza

It might be easier to understand and find the solution if you use a Venn diagram.

10. Dec 18, 2006

### dontdisturbmycircles

I understand your post quasar, thanks. I am gonna go make a coffee and then will come back and look this problem over again.

11. Dec 18, 2006

### quasar987

The word "OR" is written mathematically as the union "$A\cup B$" and the world "AND" is the intersection: "$A\cap B$".

And Poincaré's formula is just the "addition rule".

It says how you can decompose the probability of a union into a sum of probability of intersection.

The smart way of doing this problem though is through Venn's diagrams as drpizza suggested. (Have you seen those? http://en.wikipedia.org/wiki/Venn_diagram)

Last edited: Dec 18, 2006
12. Dec 18, 2006

### dontdisturbmycircles

Yea that is true that it would be a good idea to try a venn diagram, but I wanted to plug and chug with your equation and see if I can get it to match the venn diagram and then perhaps try to deepen my understanding :P :). Thanks alot btw I appreciate your help.

13. Dec 18, 2006

### quasar987

Indeed, after the vanishing terms are removed from the Poincaré "expansion", you're left with the same formula that you'd construct by inspection of the Venn diagram.

14. Dec 20, 2006

### morson

I got 0.15.

Last edited: Dec 20, 2006