Mathematics: Post your doubts here!

[mukki]

Care to explain please
3 A fair five-sided spinner has sides numbered 1, 2, 3, 4, 5. Raj spins the spinner and throws two fair
dice. He calculates his score as follows.
• If the spinner lands on an even-numbered side, Raj multiplies the two numbers showing on
the dice to get his score.
• If the spinner lands on an odd-numbered side, Raj adds the numbers showing on the dice to
get his score.
Given that Raj’s score is 12, find the probability that the spinner landed on an even-numbered side.

FIRST FIND THE PROBABILITY THAT RAJ's score is 12 with both even and odd numbers on the spinner.
Probability of getting an even number on the spinner is 2/5 and then of getting 12 is P(2,6)*2 + P(3,4)*2
So it becomes 2/5 *( P(2,6)*2 + P(3,4)*2)=2/45
now for the odd one. Probability of getting an odd number on the spinner is 3/5 but this time u will have to add two numbers to get 12 and hence p(6,6)=3/5*1/6*1/6=1/60
so P(SCORE=12) is 1/60 +2/45 =11/180
To find the conditional probability P(E|12) =2/45 devided by 11/180 = 8/11

 

[mukki]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_62.pdf
Q5 part iii HOW DO YOU FIND OUT THAT AN EVENT IS MUTUALLY EXCLUSIVE ??:/

 

[shan5674]

What is the formula for normal distribution?

V~N(np,npq) and Z= (x-μ)/σ

 

[Wanzi21]

how do you do both question?

 

[DragonCub]

ohh yeap.. thank you.. gretly appreciated ^^

You're welcome.

 

[Zishi]

(iii) To ensure at least one woman chosen, first step is to choose 1 from 3 women, which is 3C1.Then randomly choose the other 2 from the remaining 8 people. Overall it is 3C1 × 8C2 = 84

ohh yeap.. thank you.. gretly appreciated ^^

84 is NOT the correct answer. You're doing selections, not that first you're choosing one woman, then the other and so on. . . What I said in my above posts gives the correct answer, i.e 64. 9C3 - 6C3 = 64

 

[DragonCub]

http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_qp_62.pdf

Q5 part iii HOW DO YOU FIND OUT THAT AN EVENT IS MUTUALLY EXCLUSIVE ??:/

Mutually exclusive means two events cannot both happen, which can be represented by P(A&B) = P(A|B) = P(B|A) = 0.

 

[Zishi]

V~N(np,npq) and Z= (x-μ)/σ

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

 

[DragonCub]

84 is NOT the correct answer. You're doing selections, not that first you're choosing one woman, then the other and so on. . . What I said in my above posts gives the correct answer, i.e 64. 9C3 - 6C3 = 64

Yeah you're right. My answer involves some permutations.

 

[DragonCub]

View attachment 10396View attachment 10397


how do you do both question?

2(i) Let's say attending the training session is event A, and chosen for the team is event B.
You can see P(A) = 0.5, P(B|A) = 1 and P(B|A') = 0.6
So P(B) = 0.5 × 1 + 0.5 × 0.6 = 0.8
2(ii) P(A&B) = P(B|A) × P(A) = 1 × 0.5 = 0.5
P(A|B) = P(A&B) / P(B) = 0.5 / 0.8 = 0.625

 

[mukki]

Mutually exclusive means two events cannot both happen, which can be represented by P(A&B) = P(A|B) = P(B|A) = 0.

can u plz solve this for me in more detail

 

[DragonCub]

can u plz solve this for me in more detail

In the case, it is impossible to have two numbers that can both be greater than 8 and at the same time multiply to get 24. This means the possibility that the two events both happen is zero. So they are mutually exculsive.

 

[farrukh]

The position of 9 trees which are to be planted along the sides of a road is five on the north side and four on the south side.
(a) find the number of ways in which this can be done if the trees are all of different species.



(b) If the trees in (a) are planted at random, find the probability that two particular trees are next to each other on the same side of the road.



(c) if there are 3 cupressus, 4 prunus and 2 magnolias, find the number of different ways in which these could be planted assuming that the trees of the same species are identical. (d) If the tress in (c) are planted at random, find the probability that the 2 magnolias are on the opposite sides of the road.

Come on anyone please? This is a really tough one!

 

[DragonCub]

View attachment 10396View attachment 10397


how do you do both question?

7(i) Total number of choices = 12C3
number of choices with all different colours = 5C1 × 4C1 × 3C1
probability = 5C1 × 4C1 × 3C1 / 12C3 = 3/11
7(ii) Question says "exactly two" so it's taking 2 from the 4 green, and the other 1 from remaining 8 peppers.
probability = 4C2 × 8C1 / 12C3 = 12/55
7(iii) The number can be 0, 1, 2 and 3.
For 0, that's no green taken, or taking 0 from the 4 and 3 from the remaining 8, probability = 4C0 × 8C3 / 12C3 = 14/55
For 1, it's taking 1 from the 4 and 2 from remaining 8, probability = 4C1 × 8C2 / 12C3 = 28/55
For 2, we have calculated it in (ii), 12/55
For 3, the last possiblility, probability = 1 - 14/55 - 28/55 - 12/55 = 1/55 (You can also use 4C3 × 8C0 / 12C3 to obtain this value.)
In the diagram, since all the classes share the same width, the height ratio is the probability ratio.

 

[DragonCub]

Come on anyone please? This is a really tough one!

(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

 

[smartangel]

Q6. Permutations and Combinations

(i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy and Monica if each receives an odd number of mugs. [3]

The possible distributions between Lucy (L) and Monica (M) are:

...L9M1,...L7M3,...L5M5,..L3M7,..L1M9

= 10C9 + 10C7 + 10C5 + 10C3 +10C1
= 10 + 120 + 252 + 120 + 10 = 512

(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a different design. Find in how many ways these 9 mugs can be arranged in a row if the china mugs are all separated from each other. [3]

Alright, let's try this out in a slightly creative way;

Plastic Mugs ~ \_/
China Mugs ~ (_)3

If the 3 China mugs are all separated from each other, they must fit in any of the alternating 7 _ !

_ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _ \_/ _

The (_)3 have 7 _ to fit in and the \_/ can arrange in 6! ways between themselves;

n = 7P3 x 6! = 151200

(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs. These 14 mugs are placed in a row. Find how many different arrangements of the colors are possible if the red mugs are kept together. [3]

If the red mugs are kept together, the remaining 11 mugs can be arranged randomly;

\_/ \_/ \_/ x 11 !...................... [Here, we are considering the \_/ to be 'attached' so they're always kept together]

= 3! x 11 ! x 2!......................[since there are 2 possible arrangements: \_/ \_/ \_/ x 11 ! or 11! x \_/ \_/ \_/ and the \_/ can arrange in 3! ways between themselves!]

but wait! we still have to consider the 7 and 4 identical yellow and blue mugs!

= 3! x 11! x 2! = 3960......Q.E.D
........7! x 4!

DUDE. YOU'RE AMAZING. thanks a million!

 

[farrukh]

(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

thankyou so much bro! u r a genius! GOD bless u

 

[Zebedee]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_ms_63.pdf

guys, dont understand the answer for question 2. help please

 

[mominzahid]

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....

 

[mukki]

Come on anyone please? This is a really tough one!

(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