Mathematics: Post your doubts here!

[user]

plz jus explain the ans for the (iv) part. i understood the rest.View attachment 10624
here's the mark scheme:
View attachment 10625

AsSalamoAlaikum Wr Wb!
Explanation to Q:6 of Nov:2010 # 1

(iii) 4 different colors

Choose 4 different colors

and find the no. of arrangements.

6C4 x 4! = 360

(iv) Let’s say he choses 3 colors x, y, z è 6C3

Now since total of 4 pegs needed, so either x needs to be 2 or y or z so it’ll be 4!/2! + 4!/2! + 4!/2!
^this is the no. of arrangements for one choice.

Total no. of choices are 6C3

So total no. of different arrangements for 3 different colors = 6C3 x (4!/2! +4!/2! + 4!/2!) = 720

(v) any color means it could be either 4 different colors [ofc it can’t be more than that ] or 3 different colors or 2 different colors [cant be less than that, cuz we have only 2 of each color, so min should be 2 colors so as to have a total of 4]

Let’s say he chooses 2 colors x and y è 6C2

Since a total of 4 are needed, we need to have 2 of each color, and the no. of arrangement for one particular choice of colors = 4!/ (2! X 2!)

So total no. of different arrangements for 2 different colors = 6C2 x 4!/(2! X 2!) = 90

Therefore total no. of arrangement for any of her 12 pegs = 360 + 720 + 90 = 1170 ​

 

[user]

Aoa wr wb!

need help with Q:3 ii of 9709_w10_qp_61.pdf

anyone?

 

[Esme]

AsSalamoAlaikum Wr Wb!​

Explanation to Q:6 of Nov:2010 # 1​

​

(iii) 4 different colors ​

​

Choose 4 different colors ​

​

and find the no. of arrangements. ​

​

6C4 x 4! = 360​

​

(iv) Let’s say he choses 3 colors x, y, z è 6C3​

​

Now since total of 4 pegs needed, so either x needs to be 2 or y or z so it’ll be 4!/2! + 4!/2! + 4!/2! ​

^this is the no. of arrangements for one choice.​

​

Total no. of choices are 6C3​

​

So total no. of different arrangements for 3 different colors = 6C3 x (4!/2! +4!/2! + 4!/2!) = 720​

​

(v) any color means it could be either 4 different colors [ofc it can’t be more than that ] or 3 different colors or 2 different colors [cant be less than that, cuz we have only 2 of each color, so min should be 2 colors so as to have a total of 4]​

​

Let’s say he chooses 2 colors x and y è 6C2​

​

Since a total of 4 are needed, we need to have 2 of each color, and the no. of arrangement for one particular choice of colors = 4!/ (2! X 2!) ​

​

So total no. of different arrangements for 2 different colors = 6C2 x 4!/(2! X 2!) = 90​

​

Therefore total no. of arrangement for any of her 12 pegs = 360 + 720 + 90 = 1170 ​

Thank you so much !! +D

 

[mukki]

plz jus explain the ans for the (iv) part. i understood the rest.View attachment 10624
here's the mark scheme:
View attachment 10625

3 different colors mean that 1 of the 3 colors will be the same as the 4th color.
2same colors,2different colors = 6c1 *5c2 *4!/2!
6c1 is for selecting the color which is same in the holes and 5C2 is for selecting 2 colors which should be different. now as we have 2 identical colors the arrangements will be 4!/2!

 

[Mad-Eye]

Part a is either 52C8 or 52P8. If each picture is different then its P.
b) 12C3 * 20C3 * 20C2 OR P as im not sure about the pictures being different.

I got the answer for the first only.
Since the cards are replaced, the answer for the first is 52^8.

 

[leadingguy]

can any one explain me the story of being inclusive, exclusive independant???? plz???

 

[omar hazem]

ok dude listen this might calm you whenever N(number of trials) is very large like above 20 or so always do np and nq to check whether it converges to normal. mostly 5 marks qs like this are to be converted to normal. but always do check. if np >5 or nq >5 then uve got a normal

normal distribution u mean??
check this question in this paper and why this specially they used the cc??

 

[mukki]

anyone?

sorry for the late reply dude but first here you have to use normal to calculate the probability
P(x>32.4865)
26.4+1.645(3.7)=32.4865
once u have standardized ul find the probability of very slow which is 0.05 use binomial then and ur done

 

[mukki]

normal distribution u mean??
check this question in this paper and why this specially they used the cc??

whenever you are converting from binomial to normal you should use cc

 

[mukki]

normal distribution u mean??
check this question in this paper and why this specially they used the cc??

its like u have to use cc when ever this type of situation occurs whether they have mentioned suitable approximation or not

 

[hm12]

does anyone have any good notes about permutations and combination? i am really scared about this topic? especially its questions in recent past papers are very difficult :/

 

[omar hazem]

its like u have to use cc when ever this type of situation occurs whether they have mentioned suitable approximation or not

so generally i have to use cc when n is a large number and p is not too far from 0.5 so when can i know when to use the normal distribution without cc??

 

[nem0]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_62.pdf

6.i
i cant seem to construct the prob. distribution table.

 

[omar hazem]

so generally i have to use cc when n is a large number and p is not too far from 0.5 so when can i know when to use the normal distribution without cc??

btw i know what u said but i'm asking for P(X GREATER THAN OR EQUAL TO 235) like in this paper they used 234.5 instead of 235 and they didn't refer to any usage of suitable approximation or normal approximation so why they use this value??

 

[lavanyamane]

I have a couple of doubts in stats 1:
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf - Q.2(i) What do they mean by 'or both'? I can work out either female of watches the show, but what's "or both"?
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_62.pdf - Q.7(iv)

 

[mukki]

so generally i have to use cc when n is a large number and p is not too far from 0.5 so when can i know when to use the normal distribution without cc??

itl be written in the qs that X follows a normal distribution or X-N
and i didnt get what u mean by p is not too far from 0.5...............Just check if np or nq >5

 

[mominzahid]

can some please please explain the 2nd part?
how do we calculate the number of combinations for both albert and tracy to be on the committee?

 

[user]

sorry for the late reply dude but first here you have to use normal to calculate the probability
P(x>32.4865)
26.4+1.645(3.7)=32.4865
once u have standardized ul find the probability of very slow which is 0.05 use binomial then and ur done

the thing is i dont get the probability as 0.05
can u show me how?

 

[lavanyamane]

can some please please explain the 2nd part?
how do we calculate the number of combinations for both albert and tracy to be on the committee?

I'm stuck on this one too

 

[mominzahid]

I'm stuck on this one too

Im so gonna flunk tomorrows exam.. :'(
i suck at perms/combs.. :/