Probability Help PLEASE!! (3 Viewers)

[GaDaMIt]

Four basketball teams A,B,C and D each consist of 10 players numbered from 1,2,3...10. Five players are to be selected at random from the four teams. Find the probability that of the 5 five players selected

a) three are numbered 4 and two are numbered 9
i dont get why what im doing is wrong =\

that combination is only possible in one way.. and total possibilities = 40C5 doesn't it? so thats 1 / 40C5... but that seems far off from the answer =\


b) at least four are from the same team
i dunno how to do this

 

[SoulSearcher]

a) three are numbered 4 and two are numbered 9

Ok, from the 4 teams that are there, 4 people are wearing the number 4 jersey and 4 people are wearing the number 9 jersey. Therefore the amount of combinations of picking 3 people wearing the number 4 jersey are 4C3 and the combinations of picking 2 people wearing the number 9 jersey are 4C2. Therefore the probability is (4C3*4C2) / 40C5 = 24/658008 = 3/82251

 

[SoulSearcher]

b) at least four are from the same team

For this one, we have to consider the amount of combinations of 4 people coming from one team and 1 person from another team and 5 people coming from one team.

For 4 people from one team and 1 person from another team, the amount of combinations are 10C4 * 10C1, but since there are 4 teams, multiply by 4, and since there are 3 other teams the single person can be chosen from, multiply by 3, so that the total amount of combinations are 10C4 * 10C1 * 4 * 3 = 25200

For 5 people from one team, the amount of combinations that can occur is 10C5 * 4, as there are 4 teams that the 5 people can be chosen from. So 10C5 * 4 = 1008.

So total probability of this occuring is (25200 + 1008) / 40C5 = 28/703

 

[GaDaMIt]

Ok started Circle probability now.. one of the first questions on the exercise.. just to make sure im doing it correctly

Bob, Betty, Ben, Brad and Belinda are to be seated at a round table. Find how many ways this can be done:

a) if there are no restrictions ANSWER=24
b) if Bob is to sit in his favourite chair ANSWER=24
c) if Betty sits on Bob's right-hand side ANSWER=6
d) if Brad is to sit between Bob and Ben ANSWER=4
e) if Belinda and Betty are not to sit next to one another ANSWER=12

 

[webby234]

a) 4! = 24

b) 4! = 24

c) 3! = 6

d) 2! x 2 = 4

e) 4! - 3! x 2

= 24 - 12
= 12

Looks pretty good to me.

 

[icycloud]

I do not follow your reasoning, here is how I figured it, we look at all the ways that violate the conditions, those of the form of how many pairs are together:

AABBCCDD (4 pairs) = 4! = 24
AA?BB?CC (3 pairs) = 5!/2 - 24 = 36
?AA??BB? (2 pairs) = 6!/4 - 24 - 36x2 = 84
???AA??? (1 pair) = 6!/8 - 24 - 36x3 - 84x3 = 984

Then we sum each of these results by the ways they can happen using different orders of pairs = 24 + 4x36 + 6x84 + 984 = 1656, this subtracted by all the possible arrangements (8!/16 = 2520) is 864.

This probably makes absolutely no sense to anybody and is probably far more convoluted than the previous solution but I am pretty sure the question is beyond the scope of a highschool probability question.

Sorry a bit slow on this question, but isn't it just 4! * 3!² = 864?

 

[Riviet]

What's your reasoning Icycloud? =D

 

[GaDaMIt]

Ok started Circle probability now.. one of the first questions on the exercise.. just to make sure im doing it correctly

Bob, Betty, Ben, Brad and Belinda are to be seated at a round table. Find how many ways this can be done:

a) if there are no restrictions ANSWER=24
b) if Bob is to sit in his favourite chair ANSWER=24
c) if Betty sits on Bob's right-hand side ANSWER=6
d) if Brad is to sit between Bob and Ben ANSWER=4
e) if Belinda and Betty are not to sit next to one another ANSWER=12

for c) i dont get why its 3!.. like.. ok sit down bob first.. so thats 1 .. then betty can go either right of him.. or 2 places to right of him.. so then thats x 2 and then the others is just x 3 x 2 x 1

so thats 1 x 2 x 3 x 2 x 1.. someone explain to me why this isnt so?


does it mean the immediate right? coz 2 places to the right is still on the right hand side of bob if it is a 5 chair table..

 

[Riviet]

Since Betty must sit on Bob's right hand side, we fix them in one spot, leaving us with the 3 others, who can be seated in 3! ways.

 

[GaDaMIt]

Since Betty must sit on Bob's right hand side, we fix them in one spot, leaving us with the 3 others, who can be seated in 3! ways.

So is that a yes? It has to be on the immediate right hand side?

 

[Riviet]

Yep, the question wasn't well worded, it would've been better with something like "in the seat adjacent bob on his right hand side".

 

[webby234]

Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirts, 2 of each style. How many different arrangements are possible if no two identical shirts are next to one another?

Sorry a bit slow on this question, but isn't it just 4! * 3!² = 864?

So 4! arranging one of each type of shirt, then 3! arrangements of the next three shirts (would be 5P3 but each shirt has two places it can not go), then fit the last shirt in one of 6 places (two places it can not go in)?

4! x 3! x 6 = 864

Hi Aaron; ready for the test next week?

 

[webby234]

does it mean the immediate right? coz 2 places to the right is still on the right hand side of bob if it is a 5 chair table..

On a circular table everyone is to the right of you.

 

[GaDaMIt]

The letters A, E, I, P, Q and R are arranged in a circle. Find the probability that:

d) at least two vowels are next to one another

Edit: nevermind.. figured it out

 

[GaDaMIt]

A sports committee consisting of four rowers, three basketballers and two cricketers sits at a circular table.

a) How many different arrangements of the committee are possible if the rowers and basketballers both sit together in groups, but no rower sits next to a basketballer
b) One rower and one cricketer are related. If the conditions in (a) apply, what is the probability that these two members of the committee will sit next to one another?


Another question. Not a question from my textbook, just curiosity got the better of me cause my textbook doesn't seem to cover it.

There was one question that asked total possible ways of arranging 12 marbles in a circle, if 8 are blue, 3 red and 1 green. The answer to that was 11! / (8! x 3!). I figured this out by myself

However, how would you do such a question if there were 8 blue, 2 red and 2 green marbles? Cause the book didn't outline how you deal with such questions in relation to circle probability..

 

[Templar]

The four rowers can sit in 4! different orderings within themselves. The basketballers can sit in 3! different orderings, while the cricketers can swap around.

So 4!3!2! different arrangements.

The answer would be (12-1)!/(8!2!2!). Instead of the 12! in linear, it's 11! because by rotation the circular cycle you can generate 12 independent linear cases.

 

[GaDaMIt]

The answer would be (12-1)!/(8!2!2!). Instead of the 12! in linear, it's 11! because by rotation the circular cycle you can generate 12 independent linear cases.

Yes but by going off that that is to say that...

say there are 12 marbles.. all red.. how many different arrangements in a circle

(12-1)! / (12!) = something less than 1.. when the obvious answer is 1.. explain this?

 

[GaDaMIt]

No that's not true. Arrangements are different to combinations.

So you're saying there are a twelth arrangements for this particular case? This is confusing me =\

 

[Raginsheep]

Yes but by going off that that is to say that...

say there are 12 marbles.. all red.. how many different arrangements in a circle

(12-1)! / (12!) = something less than 1.. when the obvious answer is 1.. explain this?

The obvious answer is 1 because the red marbles are considered identical so theres no difference between the 1st marble and the 2nd or the 2nd and the the third and so on.

Don't know where you get the (12-1)!/12! from though.

 

[Riviet]

If the twelve marbles are identical, then no matter how you arrange them in a circle, you will get the same combination. Even if you go about fixing one, no matter how you arrange the other 11, you still get the same arrangement. Therefore it's just 1 way.