Mathematics: Post your doubts here!

[mominzahid]

(a) 5!*4! (for arranging the 5trees and 4 trees which are on opposite sides of the road)

(b) (4!*2! *3!*2!)/5!*4!
im not even sure if these answers are correct

(a) Simply 9! = 362880
(b) Regard the two trees as a unit. There are 7 locations available to place this unit. (Any two neighbouring slots can form a valid location but not on different sides.) 7 locations for 1 unit to fit in, that's 7P1.
Next, there are 7 slots left (9 minus 2) for the remaining 7 trees to fit in. This is 7P7.
In addition, in that unit, the 2 trees can be placed in different orders, so it's an extra 2P2.
Total number of ways is then 7P1 × 7P7 × 2P2 = 70560
(c) You can first plant the 2 magnolias, 1 on either side. So that is 4P1 × 5P1.
Then plant the remaining 7 trees in the 7 slots left. 7P7
It is notable that among the 7, 4 are of one species and 3 are of another. So the number should be divided by 4P4 then by 3P3.
Final answer is 4P1 × 5P1 × 7P7 / (4P4 × 3P3) = 700

I have a severe confusion in normal distribution and binomial distribution.. :'(
i mainly have problem identifying when to apply continuity in conversion from binomial to normal. and this is as i've figured out mainly because i cant seem to identify the questions in which the continuity correction is to be applied.. Can you Please PLEASE explain to me what type of question does the examiner give in which the continuity is to be applied and what type of question he gives when it doesnt has to be applied.. please explain the major differences between the two questions if possible with an example.. its maths p6 day after tomorrow and it would really mean alot if u could help me.. Thanks in advancee....​

 

[farrukh]

I have a severe confusion in normal distribution and binomial distribution.. :'(​

i mainly have problem identifying when to apply continuity in conversion from binomial to normal. and this is as i've figured out mainly because i cant seem to identify the questions in which the continuity correction is to be applied.. Can you Please PLEASE explain to me what type of question does the examiner give in which the continuity is to be applied and what type of question he gives when it doesnt has to be applied.. please explain the major differences between the two questions if possible with an example.. its maths p6 day after tomorrow and it would really mean alot if u could help me.. Thanks in advancee....​

it is applied when u are approximating binomial as a normal distribution.....the question will ask to use a binomial to normal approximation or if a very large number of trails are done.....then u should use normal distribution

 

[farrukh]

(a) 5!*4! (for arranging the 5trees and 4 trees which are on opposite sides of the road)

(b) (4!*2! *3!*2!)/5!*4!
im not even sure if these answers are correct

your answers are wrong...check DragonCub's solution...it is perfect except that he didnt calculate probabilities

 

[shan5674]

So what is the value of Z at probability of (1-0.1016)?

0.8984

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_62.pdf

4(ii) anyone?

 

[mominzahid]

it is applied when u are approximating binomial as a normal distribution.....the question will ask to use a binomial to normal approximation or if a very large number of trails are done.....then u should use normal distribution

Will the question always use the word 'approximation' where we are expected to apply this continuity correction?

 

[Zishi]

0.8984

That's the probability of the value of a number (z) in the table. Get get value from the table and use your normal distribution formula to get sigma in terms of mu. Now you'll have two simultaneous equations, so solve them to find the mean and standard deviation. If you're able to do that, tell me then - I'll guide you to how to do next part.

 

[shan5674]

That's the probability of the value of a number (z) in the table. Get get value from the table and use your normal distribution formula to get sigma in terms of mu. Now you'll have two simultaneous equations, so solve them to find the mean and standard deviation. If you're able to do that, tell me then - I'll guide you to how to do next part.

Okay got that. Thanks. What about part (ii)?

 

[Zishi]

Okay got that. Thanks. What about part (ii)?

Y ~ N(33, 21). If you're familiar with a normally distributed curve, then you'd know that P(33 − a < Y < 33 + a) = 0.5 is the area in the mid of the curve. As the max probability is 1, so the sum of areas left is also 0.5. Then what does the area greater than 33 + a and lesser than 33 − a equal to?

 

[shan5674]

Y ~ N(33, 21). If you're familiar with a normally distributed curve, then you'd know that P(33 − a < Y < 33 + a) = 0.5 is the area in the mid of the curve. As the max probability is 1, so the sum of areas left is also 0.5. Then what does the area greater than 33 + a and lesser than 33 − a equal to?

0.5? :S I don't get it :/

 

[Aarjit]

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

4(ii) anyone?

[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

 

[Zishi]

0.5? :S I don't get it :/

 

[DragonCub]

your answers are wrong...check DragonCub's solution...it is perfect except that he didnt calculate probabilities

Oops... I didn't notice the "probability" AGAIN. (I lost some marks in the Mock due to this. ) Really sorry about that dude.
For (b), the probability is 70560 / 362880 = 7/36.
And for (c)... Well I guess I skipped directly to (d) just now.
(c) doesn't need to concern about the magnolias, so answer is 9! / (4! × 3! × 2!) = 1260.
For (d) it's 700 /1260 = 5/9.
My answer was not that perfect anyway.

 

[Wanzi21]

Help me with this.. thanks

 

[shan5674]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_63.pdf

4(ii)?

 

[DragonCub]

View attachment 10436

Help me with this.. thanks

The (i) part is always the most direct in all P & C questions
14P12 = 4.36E^10
(ii) -The 3 seats on the right are taken by the 3 businessmen. So first it's 3P3.
-Then the Lins sit in a row on the same side of the aisle, the available locations could only be the 3 pairs of seats on the upper part. (Front pair excluded since taken byy the businessmen.) Moreover the couple can switch seats with each other. It's 3P1 × 2P2.
-Next are the Browns. 1 pair in the 3 taken by the Lins, so for them, 2P1 × 2P2.
-Finally, the students. You can see there are just the right number of seats left: 5 window seats for 5 students. 5P5.
Answer: 3P3 × 3P1 × 2P2 × 2P1 × 2P2 × 5P5 = 17280
(iii) If they seat randomly, we are back to (i) then. Total number of arrangements is 4.36E^10
-If to ensure that Mrs Brown in the front, there are only 3 seats for her to sit in. 3P1
-To ensure Mrs Lin behind a student, group her and one student together as a unit. (Note that she cannot switch the seats with that student since she must be behind the student.) For this "unit", there are 10 pairs of seats valid. 10P1.
--There a 5 students that can be grouped with Mrs Lin, so for the case above, it's 5 × 10P1.
-Finally the others. There are 14 - 1 -2 = 11 seats for 12 - 1 - 2 = 9 people. 11P9.
Number of arrangements = 3P1 × 5 × 10P1 × 11P9 =2.99E^9
Probability = 2.99E^9 / 4.36E^10 = 25/364 OR 0.0687

I'm not sure if my answers are correct.

 

[Wanzi21]

The (i) part is always the most direct in all P & C questions
14P12 = 4.36E^10
(ii) -The 3 seats on the right are taken by the 3 businessmen. So first it's 3P3.
-Then the Lins sit in a row on the same side of the aisle, the available locations could only be the 3 pairs of seats on the upper part. (Front pair excluded since taken byy the businessmen.) Moreover the couple can switch seats with each other. It's 3P1 × 2P2.
-Next are the Browns. 1 pair in the 3 taken by the Lins, so for them, 2P1 × 2P2.
-Finally, the students. You can see there are just the right number of seats left: 5 window seats for 5 students. 5P5.
Answer: 3P3 × 3P1 × 2P2 × 2P1 × 2P2 × 5P5 = 17280
(iii) If they seat randomly, we are back to (i) then. Total number of arrangements is 4.36E^10
-If to ensure that Mrs Brown in the front, there are only 3 seats for her to sit in. 3P1
-To ensure Mrs Lin behind a student, group her and one student together as a unit. (Note that she cannot switch the seats with that student since she must be behind the student.) For this "unit", there are 10 pairs of seats valid. 10P1.
--There a 5 students that can be grouped with Mrs Lin, so for the case above, it's 5 × 10P1.
-Finally the others. There are 14 - 1 -2 = 11 seats for 12 - 1 - 2 = 9 people. 11P9.
Number of arrangements = 3P1 × 5 × 10P1 × 11P9 =2.99E^9
Probability = 2.99E^9 / 4.36E^10 = 25/364 OR 0.0687

I'm not sure if my answers are correct.

wow all answers are correct.. a big thanks ^^ I'm not in good P & C
wonder how do u study it lol

 

[hassam]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_61.pdf
q6 part 4 .....how do i calculate midclass values

 

[DragonCub]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf

4(ii)?

If two events (say Q and R) are independent, then:
- P(Q&R) = P(Q) × P(R)
From this, we can derive that
- P(Q|R) = P(Q|R') = P(Q)
- P(R|Q) = P(R|Q') = P(R)
If any of the three is fulfilled, the two events must be independent.

A & B: If A happens, the first throw needs to be 1 for the sum to be 6: one-in-six chance. So P(B|A) = 1/6
What about P(B)? Sum 6, the set can be 1&5, 2&4, 3&3, 4&2 and 5&1. P(B) = 5/36.
P(B) ≠ P(B|A), so A & B are NOT independent.

A & C: If second is 5 (odd), first needs to be even so that product is even. So its 2, 4 or 6, three-in-six. P(C|A) = 1/2
For P(C), either one being even can make the product even (both even included). If you don't mind, you can draw up a table.

The light-blue boxes are valid sets. There are 25 sets in 36. P(C) = 25/36
P(C) ≠ P(C|A), A & C are NOT independent.

B & C: If C happens, in the 25 sets there are only 2 sets that make B happen (red boxes in the diagram).
P(C|B) = 2/25
P(B) we just calculated to be 5/36. P(B) ≠ P(C|B), B & C are NOT independent either.

NONE of the events is independent.

 

[RGBM211]

From a group of ten people,four are to be chosen to serve on a committee.
a)In how many different ways can the committee be chosen?210
b)Among the ten people there is one married couple.Find the probability that both the husband and the wife will be chosen. 2/15
c)Find the probability that the three youngest people will be chosen 1/30

can someone help me with part c) ...i have done a) and b) so no need of explanation ??????
answers in blue