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OCR MEI - S1 - 20th May 2015

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Avatar for leaf3
leaf3
1
Original post by Connorbwfc
For the last question I just did 6/20 * 5/19 which gave me 3/38.

Quite sure this is correct because the third chocolate doesn't matter

Yeah, thats what i got for it.
Why did everyone do complicated **** for the last question? It was just 6/20 x 5/19 = 3/38 ...


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Original post by Slenderman
Yeah it does. And for the hypothesis I did like

Ho: p=0.78
H1: p>0.78

where p = is blah blah

and this is the stupid bit. I did the binomial distribution of 19 and I did 1- that answer and got like 99.19% and said it was above 99% as it was a 1% sig test and did the next one to show the next below would be wrong. I got no critical region for the first one and I said that do not reject ho.

On the second one I said do not reject again as my critical region was 20... obviously I know that was wrong. **** -_-


I did that too! is that not right???
Original post by Bobhopkins05
I did that too! is that not right???


I don't know!! The second part is defo wrong. I just hope the first part with the original test I can get 8/8 If I can get that 8/8 I'm pretty sure I got a B. And i'll be really happy with that as I thought I got a D on it!!
Oh man I mucked up all the conditional probability ones.

For the combination one I divided part 2 by part 1 and got 0.311.. But I think someone else got a diff answer. Is that right?

Also ex and varx? I kept getting diff numbers on my calculator so worried I put the wrong one down.
Original post by jpetersgill
Why did everyone do complicated **** for the last question? It was just 6/20 x 5/19 = 3/38 ...


Posted from TSR Mobile


Yes, I did that but for some reason decided to put a "1-" in front of it - how many marks will I lose, 1 or 2?
Reply 166
Avatar for iamJOM
iamJOM
2
Original post by Manexopi
Oh man I mucked up all the conditional probability ones.

For the combination one I divided part 2 by part 1 and got 0.311.. But I think someone else got a diff answer. Is that right?

Also ex and varx? I kept getting diff numbers on my calculator so worried I put the wrong one down.


That combination probability should be correct.

E(X) was 4.96 and Var(X) was 1.31 to 3sf (1.309... I think)
Original post by Cadherin
Yes, I did that but for some reason decided to put a "1-" in front of it - how many marks will I lose, 1 or 2?


I did 1 minus as well
I thougbt that was right
Did anyone happen to get a Question booklet out? I'll make a mark scheme if there is.
Did people get 0.0407 or something for the P(X>=19) it was just 1- the probability from the previous answer right?
Original post by alicec11
Yeah there was no lower outliers but some upper outliers


Is it ok to say "there is an outlier at the upper end" rather than there are some outliers at the upper end?
Okay the last part of the last question needs clearing up lol... I'll write it out and try to explain my method; see if I'm right :P

Q: 20 chocolates, 6 cherry, 14 not cherry. Bob hates cherry. What is the probability of Bob getting a chocolate he likes in more than 2 picks (no replacement).

A: I thought of this as P(>2 picks to get something he likes) = 1 - P(<= 2 picks to get something he likes) = 1 - P(0 picks) - P(1 pick) - P(2 picks)

I'm not 100% sure about this now, because lots of people got different answers, however, if I'm wrong, the answer that I think is correct would be 1 - P(0 picks) - P(1 pick), but the wording of the question is really stupid to try and get your head around.

Numbers for what I did, regardless:

P = 1 - (14/20) - (6/20) * (14/19) - (6/20) * (5/19) * (14/18) = 0.009... iirc
Reply 172
Avatar for iamJOM
iamJOM
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Original post by Daniel McLovin
Did anyone happen to get a Question booklet out? I'll make a mark scheme if there is.


Yes, I'll post a picture when I'm home. It does have my answers on it too though, so you can compare those with yours.
What did we have to do for the deck of card question?
Original post by Daniel McLovin
Did anyone happen to get a Question booklet out? I'll make a mark scheme if there is.


Is there any chance you can do one for the last question? I can remember it pretty well:

20 chocolates, 7 O, 6 C, 4 X, 3 Y

First part - 3 people pick a chocolate at random, probability:
A: all pick O
B: all pick the same

Second part - P(A|B) and P(B|A)

Third part - P(A) on two seperate occasions, same situation

Fourth part - probability 1 person gets (not C) in more than 2 picks (with removal)
I was stuck of proving that P(X=6)=91/216 or something like that. But I got it in the end. Wasted time on it though
Original post by iamJOM
That combination probability should be correct.

E(X) was 4.96 and Var(X) was 1.31 to 3sf (1.309... I think)


Ok phew I got the same :biggrin:
Original post by iamJOM
Yes, I'll post a picture when I'm home. It does have my answers on it too though, so you can compare those with yours.


Did you just sneak it out of the exam lol?
Original post by JamesExtra
I was stuck of proving that P(X=6)=91/216 or something like that. But I got it in the end. Wasted time on it though


Hmm, that stumped me for a while as well - it was just 3 Binomial calculations in the end. I though it should really have been 1/2 or something, because all you needed was 1 6 regardless of the others but w/e
Reply 179
Avatar for iamJOM
iamJOM
2
Original post by Duskstar
Did you just sneak it out of the exam lol?


We've been allowed to take the question paper out of all of our maths exams so far :smile:

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