Mathematics: Post your doubts here!

[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

 

[parthrocks]

HEY CAN ANYONE PLZZZ TELL ME THAT IF THE AVERAGE NUMBER OF GETTING 5S ARE GIVEN THEN IT WILL BE THE PROBABILITY???RIGHT?MAY/JUNE 2011/62 QUESTION NUMBER 1

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
q6 part 4 .....how do i calculate midclass values

The 1st interval is about 1-100. therfore mid class value = (100+1)/2
2nd interval is 101-150, therfore (101+150)/2 = 125.5
And so on..

 

[Khan_971]

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

3 youngest to be chosen means 3C3. Which means there are 7 people left to compete for 1 place. therefore 7C1 = 7.
divide that by the total possible ways (210) to get 1/30

 

[Khan_971]

HEY CAN ANYONE PLZZZ TELL ME THAT IF THE AVERAGE NUMBER OF GETTING 5S ARE GIVEN THEN IT WILL BE THE PROBABILITY???RIGHT?MAY/JUNE 2011/62 QUESTION NUMBER 1

Could you make it clear? plus the caps lock is not recommended.

 

[DragonCub]

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

My pleasure.
There are certain tricks in solving P & C questions. The pattern is always given some subjects (people, trees, books, cards, letters etc) and slots (seats, places...), each part of the question introducing some restrictions on some of the subjects (such as which two/three letters must go together). The trick is finding out all the restrictions posed by the question and sort the groups of subjects out from the bearing most strict restrictions to being the freest.
For example, in part (ii), the business people bear the heaviest restriction (only the front seats); then the couples (same row, same side), then the students (window). I did this by allocating seats for the business people first, then the two couples followed by the students. At last I multiply them together to get the answer. This is a very safe and accurate way to solve this kind of problems.

 

[RGBM211]

3 youngest to be chosen means 3C3. Which means there are 7 people left to compete for 1 place. therefore 7C1 = 7.
divide that by the total possible ways (210) to get 1/30

still confused sorry could you explain part B as well

 

[parthrocks]

Could you make it clear? plus the caps lock is not recommended.

hey i want to ask u that if the z has a value of say 0.228 whose inverse cannot be found out from the table....thus in this case do we need to do 1-0.228 =0.772 from the value......that is something....???why bt

 

[leosco1995]

The 1st interval is about 1-100. therfore mid class value = (100+1)/2
2nd interval is 101-150, therfore (101+150)/2 = 125.5
And so on..

But why do we have to use mid-values here..?

 

[parthrocks]

hey i want to ask u that if the z has a value of say 0.228 whose inverse cannot be found out from the table....thus in this case do we need to do 1-0.228 =0.772 from the value......that is something....???why bt

anyone der??solve my doubt

 

[Khan_971]

still confused sorry could you explain part B as well

in part b. 2 people (couple) have booked their places. so assume 2 people and 2 seats are out .
leaving 2 seats to choose from 8 people. therfore 8C2 = 28. Probability of this will be 28/ALL possible combinations = 28/210 = 2/15

 

[Khan_971]

anyone der??solve my doubt

Well if you check the normal distribution. There's no value smaller than 0.5. So as in the case you just said. it has to be subtracted from 1. PROVIDED that if Z<any negative value for example.

 

[Khan_971]

But why do we have to use mid-values here..?

Well how will you calculate the mean using the table then?

 

[leosco1995]

Well how will you calculate the mean using the table then?

No, I mean why do you have to use decimal values like 50.5... I did (0+100)/2 to give 50.. and ended up with 264 instead of 268 as the answer.

 

[Khan_971]

No, I mean why do you have to use decimal values like 50.5... I did (0+100)/2 to give 50.. and ended up with 264 instead of 268 as the answer.

I assume that there cant be 0 students in a school. either way. I take at least 1 in a school. so therfore to find the midpoint (100+1)/2
try doing this way for all the columns.

 

[leosco1995]

I assume that there cant be 0 students in a school. either way. I take at least 1 in a school. so therfore to find the midpoint (100+1)/2
try doing this way for all the columns.

Hmm you are right, silly me.