Mathematics: Post your doubts here!

[smartangel]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_62.pdf
Q4 iii..plz explain

 

[Khan_971]

View attachment 10559

How about this?? is it possible to construct tree diagram?? I think it'll be big :S

LOL I think im gonna crack xD
Well yes you are right. but you cant try to see if it guarantees the correct answer.
u have to do it this way else: They are in serial, (Bag A, B, C)
P(2, not 2, 2) + P(8, Not 8, 8) + P(Not 8 , 8,8) + P(8,8, Not 8)

part ii is conditional probability, P(2,not 2,2) / the answer of (i)

iii) P(X) is the answer of (i). P(Y) calculate and see if it matches the value in (ii) or P(2, not 2, 2)

 

[Khan_971]

I donno .. I think Im still confused =[ but it helped that you tried.

And its from A2 mechanics (which Ive never studied, long story)... my paper is tomorrow and there's no one here who does A2 mech so......

M2? woah. well that really wasnt in my knowledge. Maybe just multiply V with the expression of dv/dx i showed.

 

[Khan_971]

Yupp.. you didd... i was using the same method but the answer was wrong somehow. :/
lemme try it agn and get back to u.

In case it does not. PM me with the link and number of that question.

 

[mominzahid]

meanwhile can u please explain the encircled step in the marking scheme of this question? i cant understand it.

 

[smartangel]

[Q4] H A P P I N E S S

(ii) The 9 letters of the word HAPPINESS are arranged in random order in a line. Find the probability that the 3 vowels (A, E, I) are not all next to each other. [4 marks]

Solving the problem via reverse approach;

i.e. find the number of arrangements in which A, E and I are all next to each other:

7 x A E I _ _ _ _ _ _

You may be wondering why I used the 7 x above; it's because A, E, I ('stuck' together) can alternate in 7 empty spaces between the _ . Let me number them for you:

1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7......... [in (a) A E I is in number 1]

n (A,E & I together) = 7 x 3! x 6! ............[P and S are recurring twice]
............clever, eh?.............2! x 2!

................................= 7560
now,

n (A,E & I NOT together) = # total possible arrangements - n (A,E & I together)
..........................................= [9! / (2! x 2!)] - 7560 = 90720 - 7580 = 83160

Probability of A,E & I NOT together = n (A,E & I NOT together) = 83160 = 11 .......... Q.E.D
...................................................... ..# total possible arrangements....90720....12

can you plz do part iii as well! thanks

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_62.pdf
Q4 iii..plz explain

2 Ps kicked out. 7 letters left including 2 S
in first case, lets take 1 S. (so we wont count the other S) 5 letter to fill in other 3 places. 5C3 = 10
second case, 2 S chosen. 2 places left for 5 letters. 5C2 = 10
Final = 20.

 

[Khan_971]

meanwhile can u please explain the encircled step in the marking scheme of this question? i cant understand it.

the method is :
fi(1.629)- (1-fi(1.075))
therefore fi(1.629) + fi (1.075) -1
Just as the given expression.

 

[shan5674]

ASA,​

i wanted to know when to take the mean of the data when calculating the quartiles.​

for eg. in M/J/2010 paper 61 stats,​

in question 2, we get the following steam-and-leaf diagram.​

0|2 5 6 8 8 (5)​

1|2 4 6 7 7 9 (6)​

2|1 2 3 3 3 5 6 7 (8)​

3|1 5 (2)​

now in calculating the median, we will divide the total number of data by 2 which in this case is​

21/2=10.5 so, which value will we take? 10 th value or 11thvalue or the mean of the 10th and 11th value.​

also in calculating the lower quartile, we get 21/4=5.25 here also, do we take the 5th value or the 6th value or the mean of the 5t and 6th value?​

again in calculating the upper quartile, we get 21*0.75=15.75 here again do we take the 15th value or the 16th value or the mean of the 15th and 16th value?​

sometimes in the mark scheme i have seen them taking the mean and sometimes the value immediately next to it.. im confused as to what should i do?​

WILL appreciate any help!​

Can someone please answer this? :S im confused too now :/

 

[Wanzi21]

LOL I think im gonna crack xD
Well yes you are right. but you cant try to see if it guarantees the correct answer.
u have to do it this way else: They are in serial, (Bag A, B, C)
P(2, not 2, 2) + P(8, Not 8, 8) + P(Not 8 , 8,8) + P(8,8, Not 8)

part ii is conditional probability, P(2,not 2,2) / the answer of (i)

iii) P(X) is the answer of (i). P(Y) calculate and see if it matches the value in (ii) or P(2, not 2, 2)

in MS.. why 'not 2' is 1 instead of 0/5???

 

[mominzahid]

the method is :
fi(1.629)- (1-fi(1.075))
therefore fi(1.629) + fi (1.075) -1
Just as the given expression.

But why do we subtract 1? :/

 

[angelicsuccubus]

M2? woah. well that really wasnt in my knowledge. Maybe just multiply V with the expression of dv/dx i showed.

it was technically just a math issue, so I thought you guys could help.. basically I had to show a=-1/3(v+30) and I couldn't seem to get rid of the 90

 

[Khan_971]

But why do we subtract 1? :/

U subtract the fi(1.0750 from 1. but since the sign outside is -ve, it reverses. It is a rule that P(a<Z<b) then fi(b)-fi(a)
but if P(-a<Z<b). fi(a negative value) is 1-fi(the value). the minus is kicked out. See the normal distribution table page. it is written there as well
therefore fi(b)-fi(-a) = fi(b)- [1-fi(a)] = fi(b)+ fi(a) -1

 

[Khan_971]

in MS.. why 'not 2' is 1 instead of 0/5???

0/5 is the probability of GETTING a 2 in Bag B.
1 is the probability of not getting a 2

 

[mominzahid]

U subtract the fi(1.0750 from 1. but since the sign outside is -ve, it reverses. It is a rule that P(a<Z<b) then fi(b)-fi(a)
but if P(-a<Z<b). fi(a negative value) is 1-fi(the value). the minus is kicked out. See the normal distribution table page. it is written there as well
therefore fi(b)-fi(-a) = fi(b)- [1-fi(a)] = fi(b)+ fi(a) -1

can i ask a really dumb question?
u keep referring to 'fi' what does it mean?
is it the probability?

 

[mukki]

0/5 is the probability of GETTING a 2 in Bag B.
1 is the probability of not getting a 2

man ure a robot i guess

 

[Wanzi21]

0/5 is the probability of GETTING a 2 in Bag B.
1 is the probability of not getting a 2

ohh I see.. then it helps me alot lol.. tomorrow is exam S1 btw n.n

 

[mukki]

can i ask a really dumb question?
u keep referring to 'fi' what does it mean?
is it the probability?

no dude fi is

 

[Khan_971]

that wierd symbol. ϕ is the fi. the function through which u have to go to the table

can i ask a really dumb question?
u keep referring to 'fi' what does it mean?
is it the probability?

 

[Khan_971]

ohh I see.. then it helps me alot lol.. tomorrow is exam S1 btw n.n

n.n? Yes. and day after tomorrow chem p4 as well!