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Author Topic: S1 and S2 DOUBTS HERE!!!!  (Read 63382 times)

Offline melony

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Re: S1 DOUBTS HERE!!!!
« Reply #210 on: May 25, 2011, 06:12:29 pm »
so for like 7(i) 63/ON/10 you wouldn't use it because it cant be done binomially?
thanks sooooo much you're a life saver!! - that clears that topic up for me! god bless u!!
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Offline SkyPilotage

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Re: S1 DOUBTS HERE!!!!
« Reply #211 on: May 25, 2011, 06:31:01 pm »
Quote from: melony on May 25, 2011, 06:12:29 pm
so for like 7(i) 63/ON/10 you wouldn't use it because it cant be done binomially?
thanks sooooo much you're a life saver!! - that clears that topic up for me! god bless u!!
In question 7 i) its already Normal distribution so you dont have to worry abt the C.C (0.5 thing)
I will give you a simple explanation...
If they ask about the prob. of 3 trials from 5 does this event X occur? You would use the binomial as 5C3 x ...x....
But if they ask you the prob of greater than 156 of event X occurs from 200... You could not possibly do 200C156 x...x... + 200C157 x...x... +.... Till 200C200
So thats why we use the continuity correction to approximate it as a Normal Distribution....So tht It would be much easier..So The Continuity correction here would be 15_._ You tell me  :D
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Offline melony

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Re: S1 DOUBTS HERE!!!!
« Reply #212 on: May 25, 2011, 06:57:07 pm »
156.5!!!!!!!  ;D
thank you so much! good luck tomorow!!   :)
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Offline SkyPilotage

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Re: S1 DOUBTS HERE!!!!
« Reply #213 on: May 25, 2011, 07:00:33 pm »
Quote from: melony on May 25, 2011, 06:57:07 pm
156.5!!!!!!!  ;D
thank you so much! good luck tomorow!!   :)
You are most welcome .. You tooo :D
By the way How are you in Permutations and Combinations??
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Offline Ghost Of Highbury

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Re: S1 DOUBTS HERE!!!!
« Reply #214 on: May 25, 2011, 07:10:35 pm »
Good luck for 2morrow guyz.
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Offline melony

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Re: S1 DOUBTS HERE!!!!
« Reply #215 on: May 25, 2011, 07:14:16 pm »
Quote
You are most welcome .. You tooo
By the way How are you in Permutations and Combinations??

Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard
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Offline Ghost Of Highbury

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Re: S1 DOUBTS HERE!!!!
« Reply #216 on: May 25, 2011, 07:19:00 pm »
Quote from: melony on May 25, 2011, 07:14:16 pm
Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard

True, very true. Always unsure of the answer..
*sighs* , let's just hope tomorrow's paper is easy.
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Offline SkyPilotage

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Re: S1 DOUBTS HERE!!!!
« Reply #217 on: May 25, 2011, 07:21:45 pm »
Quote from: melony on May 25, 2011, 07:14:16 pm
Not too bad, but most of the times i just get lucky, everyone, i mean EVERYONE in our class can do them but are always unsure of the answer being right, i guess it's the same with everyone else from what i've heard
Well My problems are all In permutations and combinations as they started to get HArd from 2009-2010  :'(... I hope it comes easy tmrw... Best OF luck for all Of uss!!  ;)
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Offline SkyPilotage

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Re: S1 DOUBTS HERE!!!!
« Reply #218 on: May 25, 2011, 07:37:40 pm »
June 2010 Variant 63 question 4 II)
Can some one plzzz explain the third way in the mark scheme???
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Offline melony

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Re: S1 DOUBTS HERE!!!!
« Reply #219 on: May 25, 2011, 08:12:06 pm »
Im not sure but this is how i understood it, this is the first time im looking at it

SO the first three colas are next to each other, [CCC]

lets just let that be one big C, green tea = G, oranje juice = J

there are only three different places we can place the G:

1) G C G J J    2) G C J G J    3) G C J J G

hence the 3

note that the Gs can not be moved the only thing we can move is C J J hence 3! = 4x3/2

therefore 3x3!=18 or as the mark sceme 3x(4x3/2)

hope i helped  ;), good luck everyone im out for the night!
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Offline SkyPilotage

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Re: S1 DOUBTS HERE!!!!
« Reply #220 on: May 25, 2011, 08:25:11 pm »
Quote from: melony on May 25, 2011, 08:12:06 pm
Im not sure but this is how i understood it, this is the first time im looking at it

SO the first three colas are next to each other, [CCC]

lets just let that be one big C, green tea = G, oranje juice = J

there are only three different places we can place the G:

1) G C G J J    2) G C J G J    3) G C J J G

hence the 3

note that the Gs can not be moved the only thing we can move is C J J hence 3! = 4x3/2

therefore 3x3!=18 or as the mark sceme 3x(4x3/2)

hope i helped  ;), good luck everyone im out for the night!
By the way Remember! YOu cannOT do 3! only There is a repeated term which is the 2 JJ so It must be 3!/2....Couldnt the G be put 2nd and last? so It would have more than 3 laces to be moved to..Its infact 6! SO its actually 6x3!/2....But what I dont get is why they Did 4x3/2 ?!?!
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Offline dj_dead

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Re: S1 DOUBTS HERE!!!!
« Reply #221 on: May 25, 2011, 08:40:12 pm »
Quote from: Ghost Of Highbury on May 25, 2011, 09:18:33 am
I guess the question/question paper you stated isn't the question you are confused about.

Anyway, i dont know the q, but i guess i understand ur doubt.

phi(t) = 0.7389
Meaning, for which value of 't' is the area 0.7389. For this u gotta look at the normal distribution table. You'll find it in the CIE S1 book
and the syl.

I've attached the part of the table where I found 0.7389. If you look closely, you'll see that cell (red) is in the 6th column (starting from column 'z') and the 7th row (starting from 0.0). 0.60 + 0.04 = 0.64,



ohhhhhh so DATS how u look at it...i ws lookin at it the other way round.....thankuu soooo veryyyy muchhh =D life saver i tel u :) all d bst wid ur papers bro !
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Offline Tohru Kyo Sohma

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Re: S1 DOUBTS HERE!!!!
« Reply #222 on: May 25, 2011, 11:00:05 pm »
can someone pls explain qns 3 p62 oct/nov 2010 CIE!
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Re: S1 DOUBTS HERE!!!!
« Reply #223 on: May 26, 2011, 12:55:05 am »
Quote from: mimiswift on May 25, 2011, 11:00:05 pm
can someone pls explain qns 3 p62 oct/nov 2010 CIE!


See the attachment below.

Since this is a conditional probability you find out the probability of getting a 12.

If the spinner is thrown on a even side
then there are 4 ways 12 can come.
and the probability of a spinner landing on 12 would 2/5.
So here is Probability of getting a 12 if the spinner lands on the even side
2/5 * 4/36

similarly you do this for the spinner landing on the odd side
which would be 3/5 * 1/36

Now it says it lands on a even side so the fraction should look like this
2/5*4/36
---------
(2/5*4/36)+(3/5*1/36)
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Offline Tohru Kyo Sohma

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Re: S1 DOUBTS HERE!!!!
« Reply #224 on: May 26, 2011, 01:40:26 am »
thank you so much dumb economist!
+rep
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