Probability (1 Viewer)

XcarvengerX

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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.;)
 
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CheekyPunk

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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

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

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XcarvengerX said:
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

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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

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

dunno04

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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

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silentprayer said:
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...:D
 

YBK

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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

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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!
 
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XcarvengerX

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YBK said:
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... :D
pesila said:
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 :p

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

pesila

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

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

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