
We survey 50 students on the clubs that they are in. There are 6 clubs.
club A : 8 persons
club B : 20 persons
club C : 12 persons
club D : 6 persons
club E : 3 persons
club F : 1 persons
If one random student was asked, what is the probability that this student was in at least 1 club?
hints: People in the survey don’t have to be in a club at all. At least 20 people are in a club because a person can’t be in a club twice.
Any thoughts? I think the answer is
20/50 * the chance that there are exactly 20 people participating +
21/50 * the chance that there are exactly 21 people participating +
22/50 * the chance that there are exactly 22 people participating +
.
.
.
49/50 * the chance that there are exactly 49 people participating +
50/50 * the chance that there are exactly 50 people participating
But that seems too long to be a not computational math problem.



We survey 50 students on the clubs that they are in. There are 5 clubs.
club A : 8 persons
club B : 20 persons
club C : 12 persons
club D : 6 persons
club E : 3 persons
club F : 1 persons
There are 5 clubs, but yet there are six clubs. Contradiction! By the principle of explosion, the probability of a random student being in a club is 3 giraffes.
For real though, there’s not nearly enough information to answer the question.



Originally posted by BobTheCoolGuy:
We survey 50 students on the clubs that they are in. There are 5 clubs.
club A : 8 persons
club B : 20 persons
club C : 12 persons
club D : 6 persons
club E : 3 persons
club F : 1 persons
There are 5 clubs, but yet there are six clubs. Contradiction! By the principle of explosion, the probability of a random student being in a club is 3 giraffes.
For real though, there’s not nearly enough information to answer the question.
Edited the post. And I think that is enough to solve the question.



You think so? I was thinking different students could have different likelihoods of joining clubs, but you may be right that when we just look at the probabilities for one student it doesn’t matter since it’s the average of everything.



Hmm. Let’s say we just have 2 clubs, and there are 4 students total.
Club A has 3 people and club B has 1 person.
We know the chance of a random student being in a club is at least .75.
Assuming the person in B has an equal chance of being any of the students (can we assume that?) there’s a .25 chance it’s the one student who isn’t in Club A. So, our answer would be .75*.75 + .25*1 = .8125. I think this may be the same thing you were doing up above, not sure. Also, do you know if I can assume what I did up there? Say one of our 4 students hates clubs and never would join one. Then our answer would be .75



well for a simple case like that I suppose we can have nCr(4,3) * nCr(4,1) different configurations. In all of these only nCr(4,3) * 3 of them are cases in which all but one people are participating. I don’t know where to go from there.



You’re going at it wrong. The key here is that participation in any club has no bearing over participation in any other club. In probability terms they are independent events. Therefore if you pick a random student out of the crowd there is:
 a 16% chance they are a member of club A.
 a 40% chance they are a member of club B.
 a 24% chance they are in club C.
 a 12% chance for club D.
 a 6% chance for club E.
 a 2% chance you picked the weirdo who runs the socksmelling club F by himself.
The probability that the student is not in any one club is the complementary. If we call ‘x’ the probability of a random student being in club X then the probability p of a random student being in no club is:
p = NOTA AND NOTB AND NOTC AND NOTD AND NOTE AND NOTF
p = (1 – a) * (1 – b) * (1 – c) * (1 – d) * (1 – e) * (1 – f)
p is a smidget above 31%.
p is NOT the fraction of all students who are in no club. There is an important distinction here, in that we don’t even know if there exists a student who is in no club. But, based on the information we do know, this is the most we can infer.
Also, please don’t make us do your homework qwerber.



That’s the thing, the answer the prof provided was 0.84. He worded the question in a weird way too, he’s not natively anglophone: “what is a probability that student participates in at least one club?”
At first I thought it meant what is the probability that all students are in at least a club, but that was way off.
Anyways, if I interpreted the question correctly, ace blue should be right. Just wanted to mention that the answer given was 0.84 (which is probably incorrect). Thanks.



100%, because the sort of people who answer surveys are also the sort of people who join clubs. I am more curious about club F though. What IS that?



Oops, my bad. I went for the probability that a random student is in no club. The probability that a random student is in at least one club is 1 – that, so slightly below 69%. I checked my math and stand by it. I’d be curious in learning how your teacher came to his answer.



Teachers can mistake. Yes, it seems that Ace_Blue’s math is correct.



Also, please don’t make us do your homework qwerber.
In have no problem with people posting questions like this for a couple of reasons. 1) No one has to answer. No one is making anyone do anything. 2) Even if it is their homework, what’s wrong with helping someone with a problem they need help with?
I think Ace Blue’s final answer makes sense, however, his claim that “The key here is that participation in any club has no bearing over participation in any other club” is most likely the way the problem works, but it never said that anywhere. What if Club B is the Club club that you have to be in to be in any other clubs? What if Club F is the murder’s club and people who are in it can’’t be in any other clubs, etc. What if students in the sports club have a higher chance of joining the basketball club?



Originally posted by BobTheCoolGuy:
Also, please don’t make us do your homework qwerber.
In have no problem with people posting questions like this for a couple of reasons. 1) No one has to answer. No one is making anyone do anything. 2) Even if it is their homework, what’s wrong with helping someone with a problem they need help with?
I think Ace Blue’s final answer makes sense, however, his claim that “The key here is that participation in any club has no bearing over participation in any other club” is most likely the way the problem works, but it never said that anywhere. What if Club B is the Club club that you have to be in to be in any other clubs? What if Club F is the murder’s club and people who are in it can’’t be in any other clubs, etc. What if students in the sports club have a higher chance of joining the basketball club?
The clubs are fairly equal in joinability, it’s mostly the students random preference that affects anything.



Point is that just because someone is in Club A doesn’t mean they aren’t in Club B. Nor does it mean that they are.
There are 50 students and 50 club membership cards. But if one person is in two clubs, that means that there’s someone NOT in a club.
But it doesn’t matter because the entire exercise is a sum of independent random events, thus, NOTA AND NOTB AND NOTC AND NOTD AND NOTE AND NOTF
A more difficult problem is “what is the probability that a student polled is in at least two clubs?” because now you have to factor combinations.



