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    (Original post by alicec11)
    I got Reject H1 because it was outside the significance level ?
    Hm - I got it as within the significance level...
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    Second last one I got 7/12 I think.

    For the last question, I'll write down what I remember of it:

    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)


    What I did:

    1A: (7*6*5)/(20*19*18) someone I spoke to says *3 because of order, I hope that's wrong :P

    1B: (7*6*5+6*5*4+4*3*2+3*2*1)/(20*19*18)

    2: P(AnB) = P(BnA) = P(A) because of the situation
    So P(A|B) = P(A)/P(B) whatever that was, maybe 1/19 I can't remember
    And P(B|A) = P(A)/P(A) = 1

    Next part was P(A)*P(A)

    And last part I got as:
    P(>2) = 1 - P(0) - P(1) - P(2)
    Which was 1 - 14/20 - (6/20)*(14/19) - (6/20)*(5/19)*(14/18) which was a number, I think it simplified a lot (this might have been 1/19 or w/e)

    The last question was the only one I really found hard.

    With regards to the guy asking about the hypothesis test, I got Accept H0, then reject H0 - if you look at markschemes, they don't credit you for 'accept H1' for some reason - always answer in terms of H0.
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    For the last question I did 1- ( he picks cherry twice)

    I got something like 0.079

    What did you get?
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    For the last question I've got something /57 or 1/51.. Anyone?
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    (Original post by alicec11)
    I did 6/20 * 5/19 * 15/18

    Or whatever the numbers for cherry was but I think thats how you do it
    Isn't it 6/20*5/19*(7/18+4/18+5/18)

    So that it is cherry, cherry, orange+
    cherry, cherry, coffee +...
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    how did everyone get 3/38 for the last one??


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    Last question:

    probability 1 person gets (not C) in more than 2 picks (with removal)

    So you can find out the probability of getting a cherry in one and 2.

    P(one) = 6/20
    p(two) = 14/20 * 6/19

    Add these up to get the p of getting a cherry in two picks. then i did 1 - p of this? to get the p of not?
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    Did people put 0.857 or something for the question about the deck of cards?

    And for expected it was 8.57?

    Did people put 0.036 for the combination question about what is prob that the team contains the right amount of forwards and backs?
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    For the last part of the last question, I looked at it as probability of getting something to eat in 2 or less was 1 - probability of getting something first pick - probability of getting something after 1 cherry - probability of getting something after 2 cherries
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    (Original post by Kira Yagami)
    Did people put 0.857 or something for the question about the deck of cards?

    And for expected it was 8.57?

    Did people put 0.036 for the combination question about what is prob that the team contains the right amount of forwards and backs?
    I couldn't do that forwards and backs question the last part of that question. Only bit I left blank
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    (Original post by Kira Yagami)
    Did people put 0.857 or something for the question about the deck of cards?

    And for expected it was 8.57?

    Did people put 0.036 for the combination question about what is prob that the team contains the right amount of forwards and backs?
    Yes to all of those
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    (Original post by Duskstar)
    For the last part of the last question, I looked at it as probability of getting something to eat in 2 or less was 1 - probability of getting something first pick - probability of getting something after 1 cherry - probability of getting something after 2 cherries

    That's what I did!!!
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    (Original post by JamesExtra)
    I couldn't do that forwards and backs question the last part of that question. Only bit I left blank
    It was the second part dovided by the first part - the wanted outcomes divided by the total possible outcomes
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    For the one about orange centre sweerts

    3 in a row was 0.0307

    so having this probability once in the first week and once is the second week = 0.0307 X 0.0307 right?
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    (Original post by joshgofe)
    That's what I did!!!
    Yay! My maths is at the top of page 6.
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    That was actually surprisingly a good paper
    I knew that I would struggle with the bloody binomials, so I revised them so much and literally half the paper was binomials
    Does anyone have a copy of the paper?
    What are people thinking for grade boundaries? I'd say 62


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    (Original post by Duskstar)
    For the last part of the last question, I looked at it as probability of getting something to eat in 2 or less was 1 - probability of getting something first pick - probability of getting something after 1 cherry - probability of getting something after 2 cherries
    What I did was:

    (Probability of getting cherry) x (probability of getting cherry again) x (probability of not getting cherry)

    But I don't think thats right...
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    (Original post by Kira Yagami)
    For the one about orange centre sweerts

    3 in a row was 0.0307

    so having this probability once in the first week and once is the second week = 0.0307 X 0.0307 right?
    I'm pretty sure that's right too
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    Can we leave our answers as fraction?? Or are we gonna lose marks for that??
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    (Original post by Duskstar)
    Second last one I got 7/12 I think.

    For the last question, I'll write down what I remember of it:

    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)


    What I did:

    1A: (7*6*5)/(20*19*18) someone I spoke to says *3 because of order, I hope that's wrong :P

    1B: (7*6*5+6*5*4+4*3*2+3*2*1)/(20*19*18)

    2: P(AnB) = P(BnA) = P(A) because of the situation
    So P(A|B) = P(A)/P(B) whatever that was, maybe 1/19 I can't remember
    And P(B|A) = P(A)/P(A) = 1

    Next part was P(A)*P(A)

    And last part I got as:
    P(>2) = 1 - P(0) - P(1) - P(2)
    Which was 1 - 14/20 - (6/20)*(14/19) - (6/20)*(5/19)*(14/18) which was a number, I think it simplified a lot (this might have been 1/19 or w/e)

    The last question was the only one I really found hard.

    With regards to the guy asking about the hypothesis test, I got Accept H0, then reject H0 - if you look at markschemes, they don't credit you for 'accept H1' for some reason - always answer in terms of H0.
    Ah, I was always told to never say 'reject H0', but to say either accept H1 or accept H0 - i.e. never reject a hypothesis as there is only ever some evidence to suggest either way. I think if you put 'significant' or 'not significant' followed by 'sufficient evidence' or 'insufficient evidence' and then a comment in context you pick up all the marks.

    So did you find the result as significant for the hypothesis test in the end, as mine fell within the critical region, so I accepted H1/rejected H0 in the end.
 
 
 
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