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Mathematics: Post your doubts here!

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Help how to know wats coming of the exam
karim in math every question will include every part of the syllabus..nothing will be missed in math all the chapters in the book will be involved in the questions. like literally every chapter!!
 
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Now that is where I got wrong ! Thanks again! Yup, you said it correct! Now if I made it from 1, the answer will correct. (i think) Jazaki Allah khairan!

One more thing, the perpendicular bisector and the line y= π/4 x should be dotted lines and not solid ones. The question says < not < and equal to.
 
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7i) P(white box A) = 1/6
The clip taken from box A is placed in box B BEFORE a clip is taken from box B. So as the total number of clips in box B is 9 initially, it comes 10 later.
P( red clip from Box B) = 7/10
P (W,R) = 1/6 * 7/10 = 7/60

ii) Possible probabilities (W,R) (R,R)
The first combo was calculated in part i, for the second combo as a Red clip is taken from box A and placed in box B the number of red clips in Box B is now 8, and the total no of clip is now* [Edited] 10.
P (R,R) = 5/6 * 8/10 = 40/60
total probability =7/60 +40/60 = 47/60

iii) Find the conditional probability. P(R,R) = 40/60
probability that clip from box B red = 47/60
therefore P (R|R) = (40/60)/(47/60) = 40/47

iv) Since a clip is taken from each box, the probability of taking a red clip can be 0 , 1, or 2
for O P(W,W) = 1/6 * 3/10 = 3/60
For 1-
P (W,R) = 1/6 * 7/10 = 7/60
P (R, W) = 5/6* 2/10 = 10/60
Total = 17/60
For 2-
P(R,R) = 40/60
 
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7i) P(white box A) = 1/6
The clip taken from box A is placed in box B BEFORE a clip is taken from box B. So as the total number of clips in box B is 9 initially, it comes 10 later.
P( red clip from Box B) = 7/10
P (W,R) = 1/6 * 7/10 = 7/60

ii) Possible probabilities (W,R) (R,R)
The first combo was calculated in part i, for the second combo as a Red clip is taken from box A and placed in box B the number of red clips in Box B is now 8, and the total no of clip is not 10.
P (R,R) = 5/6 * 8/10 = 40/60
total probability =7/60 +40/60 = 47/60

iii) Find the conditional probability. P(R,R) = 40/60
probability that clip from box B red = 47/60
therefore P (R|R) = (40/60)/(47/60) = 40/47

iv) Since a clip is taken from each box, the probability of taking a red clip can be 0 , 1, or 2
for O P(W,W) = 1/6 * 3/10 = 3/60
For 1-
P (W,R) = 1/6 * 7/10 = 7/60
P (R, W) = 5/6* 2/10 = 10/60
Total = 17/60

For 2-
P(R,R) = 40/60


U're fast!!!! ;) THANKS ALOT!!!:)
 
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7i) P(white box A) = 1/6
The clip taken from box A is placed in box B BEFORE a clip is taken from box B. So as the total number of clips in box B is 9 initially, it comes 10 later.
P( red clip from Box B) = 7/10
P (W,R) = 1/6 * 7/10 = 7/60

ii) Possible probabilities (W,R) (R,R)
The first combo was calculated in part i, for the second combo as a Red clip is taken from box A and placed in box B the number of red clips in Box B is now 8, and the total no of clip is not 10.
P (R,R) = 5/6 * 8/10 = 40/60
total probability =7/60 +40/60 = 47/60

iii) Find the conditional probability. P(R,R) = 40/60
probability that clip from box B red = 47/60
therefore P (R|R) = (40/60)/(47/60) = 40/47

iv) Since a clip is taken from each box, the probability of taking a red clip can be 0 , 1, or 2
for O P(W,W) = 1/6 * 3/10 = 3/60
For 1-
P (W,R) = 1/6 * 7/10 = 7/60
P (R, W) = 5/6* 2/10 = 10/60
Total = 17/60
For 2-
P(R,R) = 40/60
Do u mind explaining the second part?? If a Red clip is taken from box A and placed in box B and the number of red clips in Box B is 8, then how is the total no of clips not 10???
 
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Do u mind explaining the second part?? If a Red clip is taken from box A and placed in box B and the number of red clips in Box B is 8, then how is the total no of clips not 10???

Sorry, that was a typo, it is 10.:( I used 10 in calculations as you can see.
I meant to write that the total no. of clips is 'now' 10 not 'not' 10.
 
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One more thing, the perpendicular bisector and the line y= π/4 x should be dotted lines and not solid ones. The question says < not < and equal to.
So should it be all dotted lines?
like the vertical line equaling 1 is dotted?
and both green lines dotted also?
I really sorry for those question.
Jazaki Allah khiaran for pointing that out..! May Allah s.w.t have mercy on your family and bless you with highest grades in both worlds Aaameen!!
 
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So should it be all dotted lines?
like the vertical line equaling 1 is dotted?
and both green lines dotted also?
I really sorry for those question.
Jazaki Allah khiaran for pointing that out..! May Allah s.w.t have mercy on your family and bless you with highest grades in both worlds Aaameen!!

Yup. The should.
Seriously, it okay. I don't mind the questions. :)
 
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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_32.pdf

Q10 (iii). I have no idea how they worked out the answer and both the mark scheme as well as the er are not helpful :/

Equation of a line is y=mx+c
Here the y-intercept is zero as it is passing through the origin so c=0
Now take the equation of the curve y=x^2 e^-x
let x be as it is and m will be the equation of gradient that you found in the previous step
i.e m= e^-x(2x-x^2)
So you have
y=mx
x^2 e^-x = e^-x(2x-x^2) x
x^2 = x(2x-x^2)
x= 2x -x^2
x^2 = x
x=1
 
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Yup. The should.
Seriously, it okay. I don't mind the questions. :)
Jazaki Allah khiaran!!! Thanks alot ! Alot!

May Allah S.w.T reward you for all you help, and shower his blessing upon you and your family, and bless all of you with a happiness and grant you all Jannatul Firdous Aaameeeen!

In Sha Allah you get all A*'s ! Aameeen
 
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