Mathematics: Post your doubts here!

[panda222]

thank GOD..........ALHAMDULILAH i got some knowledge about permutations nd combinations.............but one things teasing me what to do when they ask that e.g. tell the arrangements when 5 girls are not together given there are 5 girls and 6 boys ? :unknown: :unknown:

thanks

You first find out the number of permutations when 5 girls are together and then subtract it from the total number of permutations.

That's one way. But i find it easier to do it this way:

You do 13 dashes. _ _ _ _ _ _ _ _ _ _ _ _ _

Now the guys can not be next to each other in order for girls to be apart. So....you fill in 6 of those dashes (non consecutive) with 6 'B' s.
'
Then the number of ways the boys can be arranged in those dashes is 6!.

The number of ways girls can placed in between those gaps between boys is 7 x 6 x 5 x 4 x 3. or 7P3.

The way i learnt it was using this video:

http://www.youtube.com/watch?v=zxxrR2oa ... r_embedded

Hope i hlped.

 

[NehaKush]

which question?

 

[anahita16]

HEy in need an urgent help .. P6-May-june 2008 last question..Plz Plz help me out !

(i) 1-0.75=0.25
P(any numbber other than 5)=0.25/5=0.05
P(1,5,even)=0.05*0.75*(0.05+0.05+0.05)
=0.05*0.75*0.15
=0.00563

(ii) P(5)=0.75
A binomial model can be used since probability of success is constant.
X-B(10,0.75)
P(X>/=8) = P(X=8)+P(X=9)+P(X=10)
= 10C8(0.75^8)(0.25^2)+10C9(0.75^9)(0.25)+(0.75^10)
= 0.526

(iii) Since n is large, a normal approx can be used.
X-B(n,p)
X-B(90,0.75)
X-N(np,npq)
X-N(67.5,16.875)

Applying c.c
P(X>60.5)=P(X>(60.5-67.5)/root16.875)
=P(X>-1.704)

= phi (1.704)

=0.9558

=0.956

 

[NehaKush]

june2008 p6 question 7(i)
help

 

[anahita16]

^ i did it just above.

 

[NehaKush]

ya.. ok
nov2005 p6 question 6
thnx

 

[panda222]

http://www.xtremepapers.com/CIE/Interna ... _qp_62.pdf

Question 6 1st part only.

How do you workout the probability of 1 geese and 2 geese. I can find the probability of 0 geese but no the other two. Help?

 

[IgoforA]

qp 62 will be probability of getting geese...
first you must arrange like d,d,d,d is all duck which obtain 0 geese
second is the obtaining 1 geese, d,d,d,g & d,d,g,d & d,g,d,d & g,d,d,d... there 4 prob of getting 1 geese.... soyou will do this (5/7 *4/6*3/5*2/4) * 4
then proceed with obtaining 2 geese there 6 probability... so you will get 2/7

key:
* is times

8)

 

[IgoforA]

just wanna ask a newbie question... how do you reply to someone in this forumm?? like box to box message. hah

 

[IgoforA]

http://www.xtremepapers.com/CIE/Interna ... _qp_63.pdf help me with this permutation and combination

 

[Silent Hunter]

thank GOD..........ALHAMDULILAH i got some knowledge about permutations nd combinations.............but one things teasing me what to do when they ask that e.g. tell the arrangements when 5 girls are not together given there are 5 girls and 6 boys ? :unknown: :unknown:

thanks

You first find out the number of permutations when 5 girls are together and then subtract it from the total number of permutations.

That's one way. But i find it easier to do it this way:

You do 13 dashes. _ _ _ _ _ _ _ _ _ _ _ _ _

Now the guys can not be next to each other in order for girls to be apart. So....you fill in 6 of those dashes (non consecutive) with 6 'B' s.
'
Then the number of ways the boys can be arranged in those dashes is 6!.

The number of ways girls can placed in between those gaps between boys is 7 x 6 x 5 x 4 x 3. or 7P3.

The way i learnt it was using this video:

http://www.youtube.com/watch?v=zxxrR2oa ... r_embedded

Hope i hlped.

why 13 dashes?

 

[IgoforA]

http://www.xtremepapers.com/CIE/Interna ... _ms_63.pdf q4 please help??

 

[nikhil000]

what would be p(A intersection B, if PA and PB are given...I just need to know ..Plz help me out.

 

[abcde]

what would be p(A intersection B, if PA and PB are given...I just need to know ..Plz help me out.

p (A intersection B) = p(A) + p(B) - p(A union B)

 

[nikhil000]

what if P ( A union B) is not known?

 

[abcde]

If you know the elements of A and B, you can easily figure out the union. If that's not the case, please give an example of where your problem applies.

 

[nikhil000]

Hey guys , I can solve the permutation and combination problem pretty well but at the end i always get confused whether to multiply it or add it...Plz somebody help me out with it.

 

[XPFMember]

Remember, making one particular combination...multiply them...and all the diff. combinations you got,add them!!

 

[saimaiftikhar92]

PLEASE REPLY TO THIS QUESTION SOON........BY TODAY

1-BY DIFFERENTIATING COS X / SIN X SHOW THAT IF Y= COT X THEN DY / DX = -COSEC SQAURE X.
2- BY EXPRESSING COT SQUARE X IN TERMS OF COSEC SQUARE X AND USING THE RESULT IN PART 1 SHOW THAT


INTEGRATION UPPER LIMIT 0.5 PIE AND LOWER LIMIT 0.25 PIE COT X DX= 1 - 0.25PIE


PLEASE TELL ME ASAP

 

[vaishster]

in statistics NORMAL DISTRIBUTION FORMULA (X-mu)/sigma... u hav 2 add or subtract 0.5 to the value of X depending on the sign rite ??... but in some paper mark schemes.. they dnt do it and in some they do add or subtract 0.5 to X... how can u b sure of wen 2 do this.. its really important cuz ading or subtracting 0.5 changes the whole probalbility and the answer becomz wrong.... !!!