Mathematics: Post your doubts here!

[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.

 

[leosco1995]

Another question:

In a group of 30 teenagers, 13 of the 18 males watch ‘Kops are Kids’ on television and 3 of the 12 females watch ‘Kops are Kids’.
(i) Find the probability that a person chosen at random from the group is either female or watches ‘Kops are Kids’ or both.

..does not make sense to me. P(F) = 0.4, P(W) = 16/30 and P(F or W) = (0.4 + 16/30).. this comes out to be > 1 which is obviously wrong and I don't understand the MS and even ER properly.

 

[Khan_971]

Another question:

In a group of 30 teenagers, 13 of the 18 males watch ‘Kops are Kids’ on television and 3 of the 12 females watch ‘Kops are Kids’.
(i) Find the probability that a person chosen at random from the group is either female or watches ‘Kops are Kids’ or both.

..does not make sense to me. P(F) = 0.4, P(W) = 16/30 and P(F or W) = (0.4 + 16/30).. this comes out to be > 1 which is obviously wrong and I don't understand the MS and even ER properly.

I've had same trouble. I just subtracted P(male who dont watch) from 1 to get 5/6 as the 5/6 means the prob that its a female, any gender who watches and a watching female. Only males who dont watch are left out.

If you find a way through. let me know too

 

[shanky631]

how to solve questions like these. plz help

 

[leosco1995]

I've had same trouble. I just subtracted P(male who dont watch) from 1 to get 5/6 as the 5/6 means the prob that its a female, any gender who watches and a watching female. Only males who dont watch are left out.

If you find a way through. let me know too

Hmm.. I just looked through some old notes I had, it said:

P(A or B) = P(A) + P(B) - P(A and B)

..which probably applies here because P(F) comes twice I think. Not sure, but I hope questions like these don't come in the paper.

 

[mukki]

guys if i get 74 in p1 , 46 in m1 68 in p3 how much would i need in s1 to get an A* ? the gts are between 33 and 38 in s1 plz tell cuz i feel i wont be able to make it in 40s in s1. Finding nv 10 and 2011 papers too difficult