Probabilities & Statistics 1 - OCR (not MEI) - jan 25th 2012 discussion

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My daughter did the exam and said it was OK apart from the last two questions - do you agree ?

How many marks where they worth ?

Anyone go the paper that they could share - or their answers ?

Thanks...

Reply 61

Playingwell

Do you think that was significantly harder than past papers, I think the UMS will be put up at least :P

Reply 62

Northeastgeek

Could anybody give me the anwser to the question about if 2 values of x are found what is the probability there is only one x=2 .... I got something like 0.199 but would just like to find out if thats wrong if so please tell me so i can sit and cry in the corner of my room

Reply 63

c_subs

Original post by jammy274

Question 9 you had to do P(X=3) and P(X=4). P(X=4) was just 5/9 x 4/8 x 3/7 x 2/6 = 120/3024. P(X=3) was 3 odds and 1 even so 5/9 x 4/8 x 3/7 x 4/6 x 4C3 (because the even can be in 4 places) = 960/3024
960+120=1080/ 1080/3024 = 0.357 or 5/14.

thanks at least someOne can explain the steps.
however i can't comment on the other parts u posted -cos i dont follow

Reply 64

Jedicake

Original post by c_subs

u mean the frequecy decreases, because the probability of getting 'p' is decreasing?

P(X=1) in a geometric series with p = 1/2 is naturally just 1/2.
P(X=2) = 1/2 (fail) multiplied by 1/2 (success).
P(X=3) = 1/2 * 1/2 (two fails) again multiplied by 1/2 (success).

And so on.

Thus the probability of getting it on that turn and not before decreases with every turn, so the graph had to be the top-left, reaching a bottom limit of 0.

Reply 65

dellprinter

Original post by mcquitmp

suspicious...

Reply 66

Jedicake

Original post by Northeastgeek

I got double this.

Reply 67

N1994

Original post by mcquitmp

yeah i totally agree , i found the last 2 questions really hard too.

Reply 68

Tukazoar

Original post by Northeastgeek

do you mean 0.399?

Reply 69

c_subs

Original post by MoneyOverEverythin

For bi:

i just worked out the way of getting 3odds only out of the 4, which was 5/63 x 4 because u can do it 4 ways (4C3). then you work out odds of getting exactly 4 odds so (5/9)x(4/8)x(3/7)x(2/6) then whatever that is add 20/63 and boom.

..? idk
i did som,ething like.... 4C3 x...somthing..can't remeber

Reply 70

Jedicake

Original post by mcquitmp

I agree! The last two questions were harder than usual but still manageable.

Combined the last two were about 15-20 marks.

Usually the paper itself can't be published due to either copyright constraints or the fact we're not allowed to take the paper out the exam room, but if you view the thread in its entirety you should find some of our answers.

Reply 71

N1994

Original post by N1994

yeah i totally agree , i found the last 2 questions really hard too.

questions 8 and 9 were worth 20 marks altogether.

Reply 72

c_subs

Original post by Jedicake

p[(1-p)^n-x] then?

Reply 73

mcquitmp

Original post by dellprinter

suspicious...

Well it may seem that way but I'm just a concerned parent.

If you don't trust me then fine - all you need to answer is the first bit - what marks were the last couple of questions worth so I can work out the potential impact on her score.

I've no need to cheat to get a score on these exams as I'm fairly confident with a few days prep I'd get 100% in them all if I needed to take them but with a First Class honours degree in microelectronic systems design - I'll give that a pass and not waste my time.

If you are stuck on a questions just ask me.

Reply 74

Ree69

Mr M will produce some answers hopefully

Reply 75

MoneyOverEverythin

Original post by Tukazoar

do you mean 0.399?

yeah i remember for that question. This question was the first real testing one tbh but after a good think its quite simple. Its (0.275 x 0.725)x2 = 0.399.

Reply 76

mcquitmp

Original post by Jedicake

Thanks - 20 marks is unfortunately a fair whack out of 72.

Reply 77

Jedicake

Original post by c_subs

p[(1-p)^n-x] then?

P(X=x) =

Unparseable latex formula:

p(1-p)^x^-^1

Reply 78

Playingwell

Original post by mcquitmp

Woww thats so impressive, tell us all about it.

Tw*t.

Reply 79

jammy274

Original post by c_subs

thanks at least someOne can explain the steps.
however i can't comment on the other parts u posted -cos i dont follow

Sorry, I'll try to explain better.
Last part of question 8 you have to find probability that there are more red than blue (or the other way around I can't remember). For the part before that you had to work out the probability that there are 11 of each. so you do P(R=11) 22C11 x 0.5^11 x 0.5^11. Which comes to about 0.168.
The only possibilities are that Red and Blue have the same number, there are more red, or there are more blue. Because p = 0.5 the probability that there are more red or more blue are the same. So you do 1 - 0.168, then divide by two because the other two outcomes are equally likely.

And for the last part of question 9, the only possible combinations of making 28 and 9,8,7,4 and 9,8,6,5. so to work out out the probability that it's 9,8,7,4 you do 1/9 x 1/8 x 1/7 x 1/6 = 1/3024 then x by 4! because they can be in any order. Then you do the same for 9,8,6,5 so you x by 2. That should give you 48/3024 or 0.0159