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Mathematics: Post your doubts here!

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PhyZac

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Scafalon40

Scafalon40

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PhyZac said:
well, first find "n" as follows

is the expected number is to 8 from n ,therefore, (the prob of getting a 3 ) 2/6 x n = 8
n = 24

now i hope u know this rule, that is npq is variance
n = 24
p = 2/6
q = 1 -p = 1 - (2/6) = 4/6

24 x 2/6 x 4/6 = 5.33
Click to expand...
thanks....I completely overlooked the 2/6 part....my brain is fried. I've been at this stuff since 10 am :confused:
 
Scafalon40

Scafalon40

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PhyZac said:
b (i) For getting a even number u need the number to end with even number (ofc) well in that number we have 2 4 6 8 as even,
now we consider each separtely
with 2 as end
we have 1 44 687
we rearrange 6! / 2! (we permutate 6 and we need to divide with 2 cuz there are 2 4s)
with 6 and 8 as end it is same thing as with 2
so we get 6!/2! and 6!/2!
with 4 as end we wont have 2 4s, so it will be only 6!
now add
6! + (6!/2!)*3 = 1800

i will try to solve rest soon In Sha Allah

b (ii) to get a number between 20 000 and 30 000 we have to start with number 2 and only, (we cant have a number smaller than 30000 starting with 3 ryt!)
so basically we have 2 and 4 blank
2_ _ _ _
so we permutate 4 from the left 5 numbers
we get 5P4 = 120!

(c) wat a tricky one!
anyway,
see, now since all have same prob, u get that each prob is 1/3

so we start any color and then take any of the other two in next shot
I am not sure if u got my point,
so we 1 * (2/3 )^7

first try is one, becuz all is possible..and power is 7 since 8 - 1 = 7
Click to expand...
I'm having trouble with this one too. The last part, could you explain it further please?
 
Scafalon40

Scafalon40

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PhyZac said:
Sure....hope u get it now.
View attachment 23276
Click to expand...
Ok dude that helps alot! Thanks. This is as far as I got
The spaces:
_ _ _ _ _ _ _ _
1/3 in the first slot----->got it
the next tile should not be matching, so 2/3 in the second------>got it
now the next tile after that should not match the second tile. The probabiltiy that the next tile matches the second is 1/3, and the probability that it doesn't is 2/3. So 2/3 in the third slot. And this goes on...
Point out any fault in my reasoning, if you would please.
 
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PhyZac

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Scafalon40 said:
Ok dude that helps alot! Thanks. This is as far as I got
The spaces:
_ _ _ _ _ _ _ _
1/3 in the first slot----->got it
the next tile should not be matching, so 2/7 in the second------>got it
now the next tile after that should not match the second tile. The probabiltiy that the next tile matches the second is 1/3, and the probability that it doesn't is 2/3. So 2/3 in the third slot. And this goes on...
Point out any fault in my reasoning, if you would please.
Click to expand...
Well,
In first slot it is 1 ! ( because it cud be any! no restrictions)
Now from second onwards it is 2/3 ! because a different color to first is 2/3 and so on!
We dont start with 1/3 and it is 2/3 not 2/7
 
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PhyZac said:
Well,
In first slot it is 1 ! ( because it cud be any! no restrictions)
Now from second onwards it is 2/3 ! because a different color to first is 2/3 and so on!
We dont start with 1/3 and it is 2/3 not 2/7
Click to expand...
oops corrected the 2/7, typing mistake :p
The reason I'm starting with 1/3 is because there is the probability of choosing any of the 3 tiles is 1/3. No?
 
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Scafalon40 said:
oops corrected the 2/7, typing mistake :p
The reason I'm starting with 1/3 is because there is the probability of choosing any of the 3 tiles is 1/3. No?
Click to expand...
Fine fine, now lets say
any tile would be 1/3
now the probabilities when we start with a grey as an example would be
1/3 x (2/3)^7
and when start with black it will be
1/3 x (2/3)^7
and when we start with white it will be
1/3 x (2/3)^7

so basically [3 x (1/3 x (2/3)^7)]
3 x 1/3 = 1 :p
 
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salvatore said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Please help me solve no. 3.. I'm totally confused!
Thanks

PhyZac I need urgent help :)
Click to expand...
The identities you have to know is tanx = sinx/cosx
and sin^2x + cos^2x = 1
and from the above u can also see cos^2x = 1-sin^2x

Now this is very messy, i would suggest to write what you see in paper...since it isnt neat on pc. And check Steel Arm post below...it is nicely represented !

1-tan^2(x) / 1+tan^2(x)
1-[ sin^2(x) / cos^2(x)] / 1 + [sin^2(x) / cos^2(x)]
now multiply every term with cos^2(x)
cos^2(x) - sin^2(x) / cos^2(x) + sin^2(x)
cos^2(x) - sin^2(x) / 1
(1-sin^2(x)) - sin^2(x)
1-2sin^2(x)
 
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salvatore said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_1.pdf
Please help me solve no. 3.. I'm totally confused!
Thanks
Click to expand...
ok so
1-tan^2x
1+tan^2x

this can be written as
1-((sin^2x)/(cos^2x))
1+(the same as above)

now the 1 in denominator and numerator can be written as ((cos^2x)/(cos^2x))

by using this u get that
cos^2x - sin^2x
cos^2x + sin^2x

the denominator is equal to 1 obv...so we work with numerator only
since cos^2x = 1-sin^2x....
u get 1-sin^2x-sin^2x
which is equal to
1-2sin^2x
Hence Shown!!!

Hope i helped!!!
 
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