Mathematics: Post your doubts here!

[Wanzi21]

The best way to do questions of this type is to subtract the "No woman is included" from the total number of selections that can be made.

then whats the total number? im so confused

 

[Zishi]

both :/

What is the formula for normal distribution?

then whats the total number? im so confused

Total number of selections is 9C3, 'cause you're selecting 3 people from 9 of them.

 

[DragonCub]

View attachment 10386

How do you do the part 3? permutation and combination question

(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

 

[Wanzi21]

(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 ^^

 

[ppaayas]

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.

 

[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