OCR MEI D1 15th June & M1 17th June 2016

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Reply 60

ineedA

Original post by lilrabbits

How did you get that? Thanks!

7 gave p as like 0.4 something , 6 was 0.5 something so less than 7

Reply 61

username1258398

Original post by ineedA

7 gave p as like 0.4 something , 6 was 0.5 something so less than 7

yea but the algorithm said take the first value of n for which p is less than 0.5

Reply 62

ineedA

Original post by lilrabbits

For the simulation question, how would you interpret the results? It worth like 3 marks.

what I did was ....there were 3 win 2 loss, 5 altogether 3/5 0.6, probability 0.6 2/5 0.4

Reply 63

ineedA

Original post by StrangeBanana

yea but the algorithm said take the first value of n for which p is less than 0.5

it's definitely less than 7, I might be wrong but that's what all my friends got, I can't remember the q exactly /:

Reply 64

username1204031

Original post by ineedA

it's definitely less than 7, I might be wrong but that's what all my friends got, I can't remember the q exactly /:

Dont think its less than 7. IIRC the algorithm was to find how many cards youd need to draw to have > 0.5 prob of drawing the same one twice? So if anything itd be 7 or more

Reply 65

fuzzedshadow

That paper has literally ensured I'm not going uni, there's no way I can get an A now, even with 260 ums from last year. Couldnt do the majority of the linear programming and got a total time of 14 for the activity on arc diagram

Reply 66

Wunderbarr

D1 ? I'll give MEI the D in a minute...

NotLikeThis Big.png

Reply 67

username1258398

Original post by ineedA

it's definitely less than 7, I might be wrong but that's what all my friends got, I can't remember the q exactly /:

You were finding the least number of cards you had to draw to have a 0.5 chance that two of them were the same, and there were 26 cards. You can do it with S1 probability:

Probability none are the Same with 7 cards = 1x(25/26)x(24/26)x(23/26)x(22/26)x(21/26)x(20/26) = 0.413
Probability none are the same with 6 cards = 1x(25/26)x(24/26)x(23/26)x(22/26)x(21/26) = 0.537

So the probability two are the same with 7 cards drawn is 1 - 0.413 = 0.587, but with 6 cards it's only 1 - 0.537 = 0.463. So you need 7 cards.

Reply 68

Wunderbarr

Original post by StrangeBanana

[video]https://youtu.be/cXKy_Bop0FQ[/video]

(edited 7 years ago)

Reply 69

username1258398

Original post by fuzzedshadow

you might be okay, you only need 220 ums this year so as long as you got more than 20 ums in D1 it's possible

Reply 70

KD1024

Original post by Jamesinatr

x=10, y=0, z=10 anyone?

I got exactly that, seems quite reasonable too, since y had a lot of wastage makes sense to avoid it.

Reply 71

KingAweT

Does anyone have the topic assessment answers for the MEI website?

Reply 72

Naseer Khan

The worst paper I've ever ****ing sat! And I wanted 100%...be lucky to scrape a C now

Reply 73

Re_sitter

It was 6
n=n+1 so although when trying to find out p you'd have been using 7 for the equation, you only had to select 6 cards for the probability to less than 0.5

Reply 74

bethygirl

Did you all get 48 for dijkstras ? Me and a few of my friends did but a couple got 49/50

Reply 75

Breakingglass

For the LP question I'm pretty sure when it said use the second plan with 32 cuts it meant the one you'd just created which is labelled z. Then for when they say that X>=20 that because it's cut into two pieces you have to interpret it as x>=10 and then there's three cuts for the 32cm one so Y>=26 becomes y>=8 then the sum must equal 20 and since we've eliminated the plan where you do 1x47cm 1x32cm that means x+y=20 I then plotted these on the graph and then minimised the waste. Idk if that's right but that's how I'd interpreted the poorly worded question that was that LP :frown: