Probability (1 Viewer)

[XcarvengerX]

So the paper has 9 marks on probability, but they are quite easy. These are the correct solutions.
3.(i) ⁵P₃ = 60
(ii) ⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅ = 320

6.
(i) Binomial distribution using ⁴C₃ and ⁴C₄
Answer: ⁴C₄ q⁴ + ⁴C₃ q³p
(ii) 1 - (⁴C₄ q⁴ + ⁴C₃ q³p)
= 1 - 4q³ + 4q⁴
(iii) Substituted p=1-q to ²C₁ pq + ²C₂ p²
= 1 - ²C₂ q²
= 1 - q²
(iv) P_{(team of 2 scores points)} > P_{(team of 4 scores points)}
1 - q² > 1 - (4q³ - 3q⁴)
= 1/3 < q < 1

Thanks to pesila for typing the workout, but thanks to me for all the subscripts and supscripts.

 

[ianc]

Yeah i got the same answers as you in question 3

probably question 6 as well, i can't remember

 

[CheekyPunk]

grrrrr! far out, p's and q's look too similar. i think i cubed (1-q), thinking that it was p^3
no wonder iv) wasnt working out either
><
hopefully i will still get 1 mark.

 

[Benmc]

Holy crap...i even got Question 3 wrong and didn't even do 6. :burn:

 

[pLuvia]

So the paper has 9 marks on probability, but it is quite easy. Please share your solutions to these probability questions.
3.(i) 5P3
(ii) 5P2 + 5P3 + 5P4 + 5P5 = 320 (I think)

6. (i) Binomial distribution using 4C3 and 4C4
(ii) I think it is 1 - 4q² + 4q³
(iii) Can't remember - can't be bothered to do it again
(iv) 1/3 < q < 1

Yep same, Q6 (ii) I think I didn't use the hence, just did it the long way

 

[XcarvengerX]

Yay, thanks guys, I am confident now that I got those 9 marks. Don't stress too much if you didn't get it

Still, should also get the 3 marker probability question in 4 unit paper but I didn't...

 

[silentprayer]

in 6iv) u have to develop a polynomial first right? would u get any marks for just writing that down?

 

[dunno04]

probability.
I fudge it.
OMG.
I think i've fudge the whole maths paper
NooOoOoOoooo

BUt i think i get the question 6b right

 

[XcarvengerX]

in 6iv) u have to develop a polynomial first right? would u get any marks for just writing that down?

Yeah, using part (ii) and (iii) you get a polynomial, then solve it. You have to write down how you get that as part (iv) worth 2 marks. I am just too lazy to type it here...

 

[YBK]

lol for question 3 i got the same answer but used a different method... i never really understood the P notation so I used the C thing..

i did

5! + 5C4*4! + 5C3*3! + 5C2 *2 = 320

 

[pesila]

Um okay well question 3c was 60 and 320 yeah? But question 6 I'm not too sure about all those answers. I just did it again (I probably got most of it wrong in the exam, but this is right I think)
i) binomial 4C4 q^4 + 4C3 q^3 p
ii) 1 - (4C4 q^4 + 4C3 q^3 p) works to start off since they need 2,3 or 4 to score points. The result is 1 - (4 q^3 -3 q^4)
as p = 1 - q, sub in for p to get 1 - 4 q^3 + 3 q^4
I don't think I did the whole 1- bit thinking it was probability they would lose
iii) basically 1 - 2C2 q^2 since they can score points with 1 or 2 completing, so the only way not to is if 0 complete. If you did 2C1 p q + 2C2 p^2 and substituted p=1-q that's fine, you'll get that.
iv) since it's P(team of 2 scores points) > P(team of 4 scores points) is the principle of the question,
1 - q^2 > 1 - (4 q^3 - 3 q^4)
Solving this proved to be a comparatively easy task for 2 marks given Q6aiii for 1. You get 1/3 < q < 1
A bit annoyed. I should have got that question out in about a minute for the whole deal in the exam
I think I'll lose like 3 marks on it. Oh well. So is life. Later!

PS just in case you didn't realise, but you might have YBK, you simply did nCk * k! = n! k! /((n-k)!k!) = n! / (n-k)! = nPk, so it's exactly the same! Yay me. Sorry if you did. Also I realise this is probably useless to you now, but hey, I do love my maths. Cya!

 

[XcarvengerX]

lol for question 3 i got the same answer but used a different method... i never really understood the P notation so I used the C thing..

i did

5! + 5C4*4! + 5C3*3! + 5C2 *2 = 320

That's exactly how you derive permutation formula from combination formula (by multiply it with r!). No wonder the answer is the same...

Um okay well question 3c was 60 and 320 yeah? But question 6 I'm not too sure about all those answers. I just did it again (I probably got most of it wrong in the exam, but this is right I think)
i) binomial 4C4 q^4 + 4C3 q^3 p
ii) 1 - (4C4 q^4 + 4C3 q^3 p) works to start off since they need 2,3 or 4 to score points. The result is 1 - (4 q^3 -3 q^4)
as p = 1 - q, sub in for p to get 1 - 4 q^3 + 3 q^4
I don't think I did the whole 1- bit thinking it was probability they would lose
iii) basically 1 - 2C2 q^2 since they can score points with 1 or 2 completing, so the only way not to is if 0 complete. If you did 2C1 p q + 2C2 p^2 and substituted p=1-q that's fine, you'll get that.
iv) since it's P(team of 2 scores points) > P(team of 4 scores points) is the principle of the question,
1 - q^2 > 1 - (4 q^3 -3 q^4)
Solving this proved to be a comparatively easy task for 2 marks given Q6aiii for 1
A bit annoyed. I should have got that question out in about a minute for the whole deal in the exam
I think I'll lose like 3 marks on it. Oh well. So is life. Later!

Parts (i) and (ii) and also (iii), that's what I mean. Edit first post. Sorry I typed my original post straight from memory, not by doing it again at home

Note the answer for part (iv) is derived from the inequality you gave, solve for q. So what is the problems?

 

[pesila]

Note the answer for part (iv) is derived from the inequality you gave, solve for q. So what is the problems?

I stuffed up part ii in the actual exam, which stuffed me up because I thought it was P(losing), so I got a cubic and wasn't willing to try factor theorem for the roots. With the right values for ii and iii it comes out very easily. Just clarifying that

 

[YBK]

hah.. that's cool! i derived the p formula thing without realising