Help with maths question - it's so easy I'm embarrased! (1 Viewer)

[shimmy&shine]

Ok , here it goes...

Twelve students are to be chosen from twenty students of equal ability.
Find the probabability that:

i) three particular students A, B, C are chosen

ii) students A and B are chosen but student C is not chosen

iii) none out of students A, B, or C are chosen

iv) at least one of the students A, B, or C are chosen


So many people are probably readoing this and rolling their eyes going 'that's so damn easy'. but I can't do.

Any help appreciated!

 

[SoulSearcher]

This is maybe because it looks more like a combinations question, which is a 3 unit topic. Anyway, I'll do it with combinations, but the answers might be wrong.
i) Probability of choosing A, B and C = ¹⁷C₉/²⁰C₁₂ = 11/57
ii) Probability of choosing A and B but not C = ¹⁷C₁₀/²⁰C₁₂ = 44/285
iii) Probability of not choosing A, B or C = ¹⁷C₁₂/²⁰C₁₂ = 14/285
iv) Probability of choosing at least A, B or C = 1 - ¹⁷C₁₂/²⁰C₁₂ = 1 - 14/285 = 271/285

 

[shimmy&shine]

uuhyhh, I'm really sorry buyt I only do 2 units so i have no idea what 'combinations mean', but thanks for your help. Should I know how to do this at all then?

I have no idea what all the 'C' in between the numbers mean. Is there a button on the calculators to do this?

Is there a 2U method of doing this? This was a Q8 Quezzie btw.

Soulsearcher, your answers were perfectly right. yay, but I still don't understand.

 

[SoulSearcher]

Yes, well combinations is a way of counting the number of unordered selections of distinct objects chosed from another set of distinct objects. For example, in the set {A, B, C, D}, there are 6 possible groups of 2 i.e. {A, B} {A, C} {A, D} {B, C} {B, D} {C, D}, and it doesn't matter what the order is, as it is the same group. I suspect that there is a 2 unit way of doing it, but it is not easily noticable here.

The C just represents the combinations that are possible by taking a certain number of things out of the group. If you have a Casio calculator, theres a button on the top row, second from the left that says nCr. That is the combinations button.

As I said, there is probably a 2 unit way of doing this question, but I have yet to notice it at the moment. You should not be needing to know combinations though, because it is a 3 unit topic. Ask if you need a bit more help understanding this

 

[DeathB4Life]

Ok , here it goes...

Twelve students are to be chosen from twenty students of equal ability.
Find the probabability that:

i) three particular students A, B, C are chosen

ii) students A and B are chosen but student C is not chosen

iii) none out of students A, B, or C are chosen

iv) at least one of the students A, B, or C are chosen


So many people are probably readoing this and rolling their eyes going 'that's so damn easy'. but I can't do.

Any help appreciated!

i too have no idea what all that 3U stuff is saying but heres how i would do it:

i)
Chance of A: 12/20
Chance of B: 11/19
Chance of C: 10/18

the order of being chosen shouldnt effect them so:
12/20 * 11/19 * 10/18 = ....damnit i cant find my calculator and this damn windows calculator will not suffice!

ii)
not too sure of this
Chance of A: 12/20
Chance of B: 11/19
Chance of C: 10/18
Chance of C not being chosen: 1 - 10/18 = 8/18
1/20 * 1/19 * 8/18 =

iii)
Chance of A not being chosen: 8/20
Chance of B not being chosen : 7/19
Chance of C not being chosen: 6/18
8/20 * 7/19 * 6/18 =

iv) chance of neither of atleast one of them being chosen:
1 - [chance of none of them being chosen] (the answer to Q iii)

not too sure if all (any) of that is right though.

 

[SoulSearcher]

You're right

 

[shimmy&shine]

Thank you very much Soulsearcher and Deathb4life, you maths geniuses!!!

Now that I know how to do them, it looks easy.

Your answers were spot on.

 

[SoulSearcher]

Trust me, I'm not a math genius. Flattering that you thought I was though

 

[shimmy&shine]

If it puts smile on your face, and makes you blush, lol