[Nat49]

what is the number of different ways to arrange E and A?

[aware?]

not sure :s I just did 4!+4!

[Mr M]

(Original post by Mattywooda)
Nope, he no do stats =( And yeah I got the 0.513.. just after the exam :/ spent 20mins on that xD I would say that's far harder than usual. Also a lot more discussion of this going on in the "revision" thread,

My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

[Sclarkey101]

I got 11/21 for the last one

[ILovePi]

(Original post by aware?)
difficult paper, anyone know answers to the arranging / combination questions? got 10/11 for the last one of them but I don't think it's right.

Yeh that was correct; you have 7 people to choose from originally (choosing 5, so 7C5).

Only 2 possibilities available; both J+J or neither.

If neither, one way of choosing (5C5). If both, then 2C2 x 5C3 = 10

Therefore, total = 11 ways and you're right in saying 10/11.

[ILovePi]

(Original post by Mr M)
My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

Please would you, everyone would appreciate it

[As_Dust_Dances_]

(Original post by aware?)
difficult paper, anyone know answers to the arranging / combination questions? got 10/11 for the last one of them but I don't think it's right.

I got 11/21 but I don't think I'm right neither. I'd like to see an unofficial mark scheme though!

[As_Dust_Dances_]

(Original post by Mr M)
My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

That would be great, thanks

[aware?]

(Original post by ILovePi)
Yeh that was correct; you have 7 people to choose from originally (choosing 5, so 7C5).

Only 2 possibilities available; both J+J or neither.

If neither, one way of choosing (5C5). If both, then 2C2 x 5C3 = 10

Therefore, total = 11 ways and you're right in saying 10/11.

ok thanks, thats a relief

[Mattywooda]

(Original post by Mr M)
My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

I don't think anyone would complain if you were to post them =) However, it does take a lot of time on your part :/ Generally, would you say this was a harder paper?

The problem with stats is that its harder to remember your answers as they're all obscure decimals.

[ninuzu]

Cocked up big time in this Really hoping for an A, will definitely have to resit.

[Nat49]

what was the number of different arrangements of e and a?

[ILovePi]

(Original post by Nat49)
what was the number of different arrangements of e and a?

6! = 720
2 possibilities for last one, then 6x5x4x3 for the others

[Sclarkey101]

i thought that there are 10 ways of arranging with them, but only one way without them meaning that the total number of combinations was 11, then out of 21 because thats the total number of groups. I could be wrong

[ILovePi]

(Original post by Sclarkey101)
i thought that there are 10 ways of arranging with them, but only one way without them meaning that the total number of combinations was 11, then out of 21 because thats the total number of groups. I could be wrong

the wording was GIVEN THAT either they are both chosen or neither are chosen, what is the probability they are both chosen?

Therefore, it is only the both/neither you consider. you want essentially (total ways of both)/(total ways of both+total ways of neither). Hence 10/11

[icy elemental]

(Original post by Mr M)
My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

Do you happen to have the mark scheme to the OCR not MEI D1 paper from today?

[cleverslacker]

(Original post by Mr M)
Might put them up later if there are queries.

I would absolutely love it if you did that!

[h2shin]

(Original post by Mr M)
My colleague Mr K has done some answers though and I photocopied them. Might put them up later if there are queries.

Hello, I just wanted an opinion.

In a question basically it was P(x=1) + P(x = 3)

But I thought time was 1:30 in the afternoon not in the morning so got P(x = 27) if the working is all correct will this get a mark? or is it just too completely wrong? it's out of 3.

[Joey15]

Many things got me - I hate Stats and always will. I got the Sum to Infinity=1 for the last question.

a=p=(1/20)

r=(19/20)

therefore... 1-r= 1/20

(1/20)/(1/20)=1

Not sure if that's correct. Probably wrong...

[ILovePi]

(Original post by Joey15)
Many things got me - I hate Stats and always will. I got the Sum to Infinity=1 for the last question.

a=p=(1/20)

r=(19/20)

therefore... 1-r= 1/20

(1/20)/(1/20)=1

Not sure if that's correct. Probably wrong...

Sorry, it's wrong. r is (19/20)^2. That's because you wan't every other time
Rest of the working is correct, so should gain M1