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Mathematics: Post your doubts here!

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PhyZac

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Alice123 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf
no 3ii n 5iii . thnx in advance Dug PhyZac syed1995
Click to expand...
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84

126 + 84 = 210
 
Alice123

Alice123

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PhyZac said:
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84

126 + 84 = 210
Click to expand...
very clearly explained. thanks a lot
 
whiz-kidd

whiz-kidd

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can anybody help me with this question for mechanics (M2)

O/N 2009 QP 52
Question 2 and 5

thanks!
 
Alice123

Alice123

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PhyZac said:
For question 5iii


there are 2 restrictions.

Either both question 1 and 2 of part A or None

If we consider 1st restriction.
Then we take question 1 and 2, and now 4 questions are left (we need total of 6)
No. of questions are 11 take the first 2 we get 9 left to choose 4 from.
therefore
9C4 = 126

Then if we consider 2nd restriction
Then we have to choose 6 from 9 questions ( since we arent allowed with question 1 and 2)
so 9C6 = 84


126 + 84 = 210
Click to expand...


 
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PhyZac

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Alice123 said:
Click to expand...
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
 
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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?
 
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