Probability, 92 (1 Viewer)

[ravdawg]

Hey guys I've been having a little problem with my perms and combs, any help would be greatly appreciated.

6 b)

A total of five players is selected at random from four sporting teams. Each of the teams consists of ten players numbered from 1-10.

i) What is the probability that of the five selected playerse, three are numbered 6 and two are numbered 8

ii) What is the probability that the five selected players contain at least four players from teh same team?

As i said before i dont like perms and combs, and whenever i see probability i try and use binomial whenever possible, even though it often isn't the way the question is made and makese it a lot more complex.

i) In the first part i used 'trinomial expansion' because there are 3 different possible desirealbe results, p= no of '6's (1/10) q= no of '8's (1/10) and r= the rest (4/5)

so i got (p+q+r)^5= (P+q)^5 +5 (p+q)4 r+...

However the terms with 'r' in it is irrelevent for the question so i only look at the first term

(p+q)^5= p^5+ 5 p^4q+ 10 p^3 q^2 +...

the bold term is the one required in the qusetion so i used it and got

Probability= 10 (0.1)^3 (0.1) ^2, which wasn't the answer

i know this is probably a crap method, but i wanna know why it was wrong

ii) For part two i thought i saw binomial again, where p= prob person is selected from particular team (4/5) and q= prob person is not selected in particular team (1/5)

so probability= 4C1 (1/5) ^4 (4/5) + (1/5) ^5

This too was wrong

Someone please tell me why my method doesnt work!!

Cheers

 

[zeek]

I haven't gone through your method yet but i will after this...
here's my solution.. hopefully its correct.

Initially, we have 40 players all together so the way we choose 5 players is ⁴⁰C₅. We don't do 5.¹⁰C₁ because that means we're selecting 1 player from one team in 5 ways.

i)Now moving onto the question... It asks for 3 players that are numbered 6. The question says that there are 10 players in each team and one player is a number 6 .: if there are four teams, there must be 4 players you can choose from that are numbered 6. To choose 3 number 6 players you go ⁴C₃
This means that from the 4 players we are choosing 3.

Now we also need to select 2 number 8 players. Again, we have 4 players that are numbered 8 so to choose 2 players from these 4 we do... ⁴C₂

Therefore, the total number of ways you can select 2 number 8s and 3 number 6s is ⁴C₃.⁴C₂. Hence, the probability of this happening is (⁴C₃.⁴C₂)/(⁴⁰C₅)

ii) This question has a keyword, at least four players, which means that you can choose 4 OR 5 and this must be included in your answer. To do this, first you find the number of ways to select 4 players from the team and then the number of ways to select 5 players and add them together.
The question also says from one team so that means we are only going to be choosing from 10 players.
.: Choosing 4 players from one team = ¹⁰C₄.³⁰C₁
Choosing 5 players from one team = ¹⁰C₅
.: Choosing at least 4 players = ¹⁰C₄.¹⁰C₅

.(at least 5) = (¹⁰C₄.³⁰C₁+¹⁰C₅)/(⁴⁰C₅)

 

[ravdawg]

I agree with the first one, but with the second, don't we add instead of multiply because of the 'or'.

As i said, now that i see the solution i get it, but i just can't replicate it, heaps annoying!

 

[zeek]

About your binomial probability... i think the problem lies with the fact that you're probably incorrectly calculating the values for (p+r) and q. I don't see a problem with you using binomial probability for something that has 3 cases though


EDIT:
Definition from wikipedia
The probability of an event can be expressed as a binomial probability if its outcomes can be broken down into two probabilities p and q, where p and q are complementary (i.e. p+q=1)

I think you might have to get the probability of p+q occuring, which is done through the combinations way.

Probability= 10 (0.1)^3 (0.1) ^2, which wasn't the answer

I don't really understand what you've done in this line...

 

[zeek]

I agree with the first one, but with the second, don't we add instead of multiply because of the 'or'.

As i said, now that i see the solution i get it, but i just can't replicate it, heaps annoying!

For the second one you don't add because the events are related.

Don't worry about permutations and combinations that much lol. I had trouble with them all the way through to year 12 and i just started to understand them properly after the half yearlies after having looked at a load of examples. You'll get the hang of it if you just keep making errors and understanding why.

EDIT: WOOPS i made a big mistake yes, you're write you do add because the events are not related. You can't choose 4 players from the same team at once and then choose another 5 coz then u'd have 9 players. Sorry.

 

[ravdawg]

Ohk thanks heaps Zeek, Damn perms and combs, no matter how many i do i cant get the hang of them. I've see the 'trinomial' method used in a 4u question before and it made sense so i couldnt see y i couldn't use it in this case....

 

[ravdawg]

If anyone else have any comments or thoughts on where i went wrong or why the method doesn't work please reply.

 

[followme]

for question 2, how about (4C1*10C4*30C1+4C1*10C5)/40C5 ?

isnt P(at least 4 from the same team)= P(4 from same team)+P(all from same team) ?

sry if im wrong. im so bad in probability..

hav u got the answer ravdawg?

 

[ravdawg]

the answer is28/703, and i think you are right.

 

[SoulSearcher]

Calculator says that followme is right.

 

[zeek]

for question 2, how about (4C1*10C4*30C1+4C1*10C5)/40C5 ?

i'm not sure about the 4C1 parts, but i do agree with everything else.

 

[BIRUNI]

answer:This is not binomial probability; it is a comination of harder 2 unit probabilities and selections in 3 unit.

you have 4 team and each team contains 10 player and so you have 40 players. you are going to select 5 players so you can do this in 40c5 ways.

i)you have 4 number 6 and 4 number 8 so the answer is (4c3*4c2)/40c5
ii) in this case you need to consider two cases
case1: 4 players are from the same team and the other from the remaining 30 so that can be done in 4*10c4*30c1
case2: 5 players are selected from the same team this can be done in 4*10c5

now the probability is (4*10c4*30c1+4*10c5)/40c5

 

[followme]

i'm not sure about the 4C1 parts,

choose one team from the four teams where the 4 players are chosen from?
That's what i hate about perm and comb+ probability, always shooting in the dark...

 

[ravdawg]

answer:This is not binomial probability; it is a comination of harder 2 unit probabilities and selections in 3 unit.


case2: 5 players are selected from the same team this can be done in 4*10c5

But isn't it possilbe to use binomial? And if not, why?

 

[speedie]

usuc ravin. i cant believe u couldnt get part i. o m g

 

[speedie]

ok this is how it works: (for part ii)

Probability(at least 4 from 1 team)= P(4 from 1 team) + P(5 from 1 team)
P(4 from 1 team) = 10C4 x 30C1 x 4 / 40C5
explanation: 10C4 ways of picking 4 people from one team times 30C1 ways of picking any one person from the remaining 30 people who are not on that team. But there are 4 teams that you can pick the 4 from, so x4.
P(5 from 1 team) = 10C5 x 30C0 x 4 / 40C5
explanation: 10C5 ways of picking 5 people from 1 team, and you dont pick anyone from the remaining 30 so x30C1(negligible). There are 4 teams so again, x4.

Therefore Probability(at least 4 from 1 team)= (4x10C4x30C1 + 4x10C5x30C0) / 40C5
which = 28/703

and rav, stop trying to make the problem harder than it is by using binomial prob. come to the library 2moro (thurs) and i'll teach u probability. CYL is coming too