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    I definitely disagree- give me C2 anyday
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    (Original post by usycool1)
    Either is fine.



    Drawing a sketch for these is always very helpful.

    P(Z < a) = 0.0170

    Using symmetry of the curve:

    P(Z > z) = 0.0170

    P(Z < z) = 1 - 0.0170 = 0.9380

    So z = 2.12.

    Because P(Z < a) is less than 0.5, a will be negative.

    So a = -2.12.

    Make sense?
    That makes sense now! thank you so much!!
    I just need a bit more practice with these. Thanks again.
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    (Original post by lebron_23)
    Okay, I'll try my best to explain it.

    The question looks odd because its just different notation, but essentially, P(X_1+X_2=4)=0.1 just means that the probability of getting a score of 4 from 2 rolls of the dice, or whatever it is, is 0.1. That is to say that the chance of you getting a combination of x that gives you 4 is equal to 0.1. So that means that to get a 4 you need either a 1 and a 3; a 3 and a 1; or a 2 and a 2. Soooo, once you've got that information, its a simple case of multiplying the probabilities and then adding them together (remember its multiplying because its AND, and its adding because its OR). So you end up with, P(1and3) + P(3and1) + P(2and2), which, in terms of numbers, is 2(0.1x0.3) + (0.2 0.2) which is equal to 0.1.

    For the next part, you just have to use knowledge from what you've just done. So if you look at 5, its basically asking you what is the probability that you will get a combination that adds up to 5. So you'd do P(2,3), P(3,2) and P(1,4), P(4,1) which should give 0.2. Then you do the P(4,4), which is the only combination to give 8, which should give you 0.16.

    Finally, for the last part, its similar to normal questions when you're asked to find probability of something between something else - you just do P(2) + P(3) because those are the only integer values in that range. That should then give you 0.05.

    Hope that helped!
    You're an absolute legend. Thank you massively
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    (Original post by salini1995)
    U guys can u tell me the hardest s1 paper please becoz I want to wanna try it out
    Thank u
    June 2002
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    (Original post by Fortitude)
    I kinds use the one in the book but my own style it just looks much simpler!

    EDIT: The lines are missing, great
    too simple that I cant even see whats going on lol, I get the idea though
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    On a venn diagram, how do you represent

    (A'UB')
    (A'UB)
    (AnB)'
    (AuB)'
    (A'nB')
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    Amydx6- ive got this far- 5/12= p(b) -(1/4)p(b)

    Can you treat it like normal algebra as i have got confused on how to rearrange it- Even simple maths confuses me in stats
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    Name:  Q5.JPG
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    Why is it that in the mark scheme it says everything divided by 56? isnt it meant to be (56-1)???
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    (Original post by Salvatore8)
    Name:  Q5.JPG
Views: 102
Size:  48.1 KB

    Why is it that in the mark scheme it says everything divided by 56? isnt it meant to be (56-1)???
    Why would you minus 1? 56 is the frequency

    Edit: look carefully. The formula shown is different to the actual formula used
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    (Original post by simonb451)
    Amydx6- ive got this far- 5/12= p(b) -(1/4)p(b)

    Can you treat it like normal algebra as i have got confused on how to rearrange it- Even simple maths confuses me in stats
    Yer literally just treat it like normal algebra I so it's 5/12 = 3/4 p(b) and then dive both sides by 3/4


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    (Original post by Salvatore8)
    too simple that I cant even see whats going on lol, I get the idea though
    Ha ha yeah, I know, basically the 2,7,? & 3 are the differences
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    (Original post by simonb451)
    Amydx6- ive got this far- 5/12= p(b) -(1/4)p(b)

    Can you treat it like normal algebra as i have got confused on how to rearrange it- Even simple maths confuses me in stats
    Imagine theres a 1 infront of th P(B)

    5/12 = 1xP(B) - 1/4xP(B) so 1 - 1/4 = 3/4

    5/12 = 3/4xP(B)

    then

    5/12 divided by 3/4 = P(B)

    P(B) = 5/9

    I think that should be right
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    (Original post by 3.1415927)
    On a venn diagram, how do you represent

    (A'UB')
    (A'UB)
    (AnB)'
    (AuB)'
    (A'nB')
    Search s1 Venn diagram shading on YouTube and the video by maths247 is really good for this


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    (Original post by B-Stacks)
    Why would you minus 1? 56 is the frequency

    Edit: look carefully. The formula shown is different to the actual formula used
    so the formula in the top right is wrong? bcz thats what ive been using
    if so, thanks!
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    (Original post by Fortitude)
    Ha ha yeah, I know, basically the 2,7,? & 3 are the differences
    Ahh right i get it
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    Could someone please explain how Part e and f of question 6 on the s1 jan 2011 paper is done? I don't understand this.

    Also, what topic within discrete random variables is this about as I'd like to get more practice on these type of questions.

    Thank You
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  1. File Type: pdf 6683_01_rms_20110309.pdf (167.6 KB, 33 views)
  2. File Type: pdf 6683_01_que_20110114.pdf (98.5 KB, 39 views)
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    (Original post by 3.1415927)
    On a venn diagram, how do you represent

    (A'UB')
    (A'UB)
    (AnB)'
    (AuB)'
    (A'nB')
    (A'UB')

    Everything except the intersection..

    (A'UB)

    Everything in B including intersection so P(B)

    (AnB)'

    Same as (A'UB') - everything except intersection

    (AuB)'

    Everything outside the circles

    (A'nB')

    Same as (AuB)' - everything outside the circles
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    (Original post by starfish232)
    Could someone please explain how Part e and f of question 6 on the s1 jan 2011 paper is done? I don't understand this.

    Also, what topic within discrete random variables is this about as I'd like to get more practice on these type of questions.

    Thank You
    Hi, I've seen this before, check out pages 5-6 on this thread

    EDIT: on page 6
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    (Original post by Salvatore8)
    so the formula in the top right is wrong? bcz thats what ive been using
    if so, thanks!
    It's not wrong, it's just a different formula to calculate it. If you've been getting the right answers with it then fine, however that way is not really endorsed by edexcel and in the mark schemes. Attached is the one I use and is used in the mark scheme. Name:  ImageUploadedByStudent Room1368645291.222016.jpg
Views: 150
Size:  6.0 KB
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    (Original post by B-Stacks)
    It's not wrong, it's just a different formula to calculate it. If you've been getting the right answers with it then fine, however that way is not really endorsed by edexcel and in the mark schemes. Attached is the one I use and is used in the mark scheme. Name:  ImageUploadedByStudent Room1368645291.222016.jpg
Views: 150
Size:  6.0 KB
    Thanks for that!
 
 
 
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