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OCR MEI - S1 - 20th May 2015 watch

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    oh-my cherry answer for the more than 2 picks was different
    I did:
    1-(14/20x (6/20 x 1/19)
    which is 0.845

    idk is that right? would i get any marks... or is that completely wrong
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    (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.
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    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???
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    (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!!
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    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.
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    (Original post by jpetersgill)
    Why did everyone do complicated **** for the last question? It was just 6/20 x 5/19 = 3/38 ...


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    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?
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    (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)
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    (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
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    Did anyone happen to get a Question booklet out? I'll make a mark scheme if there is.
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    Did people get 0.0407 or something for the P(X>=19) it was just 1- the probability from the previous answer right?
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    (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?
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    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
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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.
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    What did we have to do for the deck of card question?
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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.
    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)
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    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
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    (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?
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    (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
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    (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
 
 
 
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