IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: preity on May 21, 2009, 09:23:46 am
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guys help me out....ok da question is like this
In a group of 50 people,25 own a car,15 own a bicycle and 5 own both a car and a bicycle.If a person is chosen at random,find the probability that they:
a)own either a car or a bicycle but not both(how to answer question with the word BUT NOT BOTH)
b)own a bicyle given that they own a car.....
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40 altogether own car bike or both and of these 5 own both so 35 own only car or bike
35/50
B)5/15 since 5 of the 15 that own a bike also own a car
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I meant 5/25 for b since 5 out the 25 that own a car also own a bike
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thanks ya.... ;D
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heyy how do v do the followng question??
In country A 30% of people who drink tea have sugar in it. In country B 65% of people who drink
tea have sugar in it. There are 3 million people in country A who drink tea and 12 million people in
country B who drink tea. A person is chosen at random from these 15 million people.
(i) Find the probability that the person chosen is from country A. [1]
(ii) Find the probability that the person chosen does not have sugar in their tea. [2]
(iii) Given that the person chosen does not have sugar in their tea, find the probability that the person
is from country B.
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(1) 3 million people from country A drinking tea.
12 million people from country B drinking tea.
There are 15 million people from country A and B drinking tea.
P ( a person chosen from 15 million who come from country A)
= n ( People from country A who drink tea )
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n ( People from both countries A And B who drink tea )
= 3 millions / 15 millions = 1/5
(2) P(The person chosen does not have sugar in their tea )
= n(person from A drinking tea without sugar) + n(person from B drinking tea without sugar)
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n (person from both countries A and B who drink tea)
= (3X 0.7 ) + (12 X 0.35)
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15
= 2.1 + 4.2
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15
=0.42
(3) This is the conditional probability
P( The person is from country B | The person does not have sugar in their tea )
= (12 x 0.35)/ 15
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0.42
= 0.667
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Hey!
do u ppl have any link/resource..for | nCr, nPr, and arrangements | ...where i can understand it better...the pastpaper questions are confusingg!! i really need help and practice on that chapter.....the exam is on wed! :-[
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Candy...mayb u should try this website...http://www.s-cool.co.uk/alevel/maths.html
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I meant 5/25 for b since 5 out the 25 that own a car also own a bike
i dont think wut u did was right
cz 30 own a car(they dint tell that they own only ca
so 5/30
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Thanks for the link priety! :)
I have a question!! :-[
Six men and three women are standing in a supermarket queue.
(i) How many possible arrangements are there if there are no restrictions on order?
(ii) How many possible arrangements are there if no two of the women are standing next to each
other?
(iii) Three of the people in the queue are chosen to take part in a customer survey. How many different
choices are possible if at least one woman must be included?
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Solution :
(1) 6 men and 3 women are standing in a queue.
So, there are 9 persons standing in a queue.
If there are no restrictions on order,
The possible arrangement is 9 ! = 362880
(2) 2 women can not be standing together while man has no restriction.
__M1__M2__M3__M4__M5__M6__ ( Note : __are for women W1 W2 W3)
The places meant for men are fixed, they can only change among themselves.
Possible arrangement for men = 6!
Now there are 7 positions for women. Surely the women cannot take all. There
are 3 women only. They can only take 3 only.
Possible arrangement for women = 7 P 3
All together the possible arrangement = 6 ! x 7P3 = 151 200
(3) At least 1 women means there can be 1 or 2 or 3 women.
Since the survey involves choice, we use combination.
If 1 woman is included, the combination for women only is 3C1 and the men left
for 2 places only. So the combination for man is 6 C 2 .
Total combination =3C1 x 6C2 = 45
It follows that if 2 women are included the man left for 1 place only.
So the combination = 3C2 x 6C1 = 18
If 3 women are included, man has no place at all.
So the combination is 3C3 x 6C0 = 1
Total combination is 45 + 18 + 1 + 64
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me @ me@ me@,
Ans by Astar : I meant 5/25 for b since 5 out the 25 that own a car also own a bike
Ans by me@me@me : i dont think wut u did was right
cz 30 own a car(they dint tell that they own only ca
so 5/30
Ans by Astar is correct. In Statistics,
No of persons owning cars
=No of persons owning cars only + No of persons owning both cars and bicycles
=20 + 5 = 25
Please refer to the venn diagram in my attachment.
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thanks a lot louis... :)
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(2) 2 women can not be standing together while man has no restriction.
__M1__M2__M3__M4__M5__M6__ ( Note : __are for women W1 W2 W3)
The places meant for men are fixed, they can only change among themselves.
Possible arrangement for men = 6!
Now there are 7 positions for women. Surely the women cannot take all. There
are 3 women only. They can only take 3 only.
Possible arrangement for women = 7 P 3
All together the possible arrangement = 6 ! x 7P3 = 151 200
i dont get this part :( :(
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Ok so do u get that Possible arrangement for men = 6! ???
basically it means the 'position' of men could be:
__M6__M5__M3__M4__M1__M2__
__M5__M1__M2__M4__M3__M6__ and so on .....
So there are 6! positions (permutations) of men in that group..
Then do u get Possible arrangement for women = 7 P 3 ???
Basically there are 7 spaces for W1,W2 or W3. So (say )we put W2 in any 7 spaces...then we put W1 in the remaining 6 spaces (since 1 is already taken up by W2) and then we put W3 in the remaining 5 places....
So there are 7*6*5 = 7 P 3 ways of putting the 3 women in the spaces...
thats why : All together the possible arrangement = 6 ! x 7P3 = 6! x 7 x 6 x 5 = 151 200
This method is effective as it ensures that 2 women don't stand together! In fact if u check the CIE stats 1 book then it has this method in the P and C chapter!!
But is there some other method??? ???
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Where did the answer by louis and mac for my question go??? ??? ???
u guys do we need to know this formula??
Median= lower class boundary +( 0.5n-c.f of class below/class frequency) x class width
is in the syllabus?? i didnt come accross any question in the pastpapers so far...but this is formula is stated in the book i have used!
do i need to know it????? :-[ :-[
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no...,i don't think so we need that formula in exams... ;)
but it is always good to have knowledge about it :D
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hey u guys should have a look at: http://www.schoolworkout.co.uk/a_level.htm (http://www.schoolworkout.co.uk/a_level.htm)
its for another uk board but its helpful
I came across this example from some site (not the above 1). Isn't there something wrong in part (b)?? ???
Example C3: In a club with 8 males and 11 female members, how many 5-member committees can be chosen that have
(a) 4 females?
Solution: Since order doesn’t matter, we will be using our combinations formula.
How many ways can the 4 females be chosen from the 11 females in the club?
C(11, 4) = 11!/[(11 – 4)! ?4!] = 11!/(7! ?4!) = (11?10?9?8)/ (4?3?2?1) = 330 ways.
But this gives us only the 4 female members for our 5-member committee. For a 5-member committee with 4 females, how many males must be chosen? Only 1. How many ways can the 1 male be chosen from the 8 males in the club?
C(8, 1) = 8!/[(8 – 1)! ?1!] = 8!/(7! ?1!) = 8 ways.
Thus we have 8?330 = 2640 ways to select a 5-member committee with exactly 4
female members from this club.
(b) At least 4 females?
Solution: In part (a) we found that there are 2640 ways to select exactly 4 females for our 5-member committee. If we need at least 4 females, we could have 4 females or 5 females, couldn’t we? How many ways can we select 5 females from the 11 females in the club, given that order doesn’t matter?
C(11, 5) = 11!/[(11 – 5)! ?5!] = 11!/(6! ?5!) = (11?10?9?8?7)/ (5?4?3?2?1) = 462 ways.
So to have at least 4 females, we could have C(11, 4) ?C(8, 1) or C(11, 5). Since there are 330 ways to achieve the first and 462 ways to achieve the second, we have
330 + 462 = 792 ways
to select a 5-member committee with at least 4 female members from the club.
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yea in this question they took 330 instead of 2640. the correct answer will be:
2640+462=3102
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can someone please explain me the answer for this question..and why that method is used to solve! :'( :'(
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can someone please explain me the answer for this question..and why that method is used to solve! :'( :'(
for this question we first find out the probabilities of the two groups seperately and multiply them. we divide each group's arrangement by the number of similar objects it has, so:
for the first group:
6!/4!2! = 15
for the second group:
6!/2!3! = 60
15x60=900
there are 900 possible arrangements
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Thanx a lot ocean!! :D :D
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how do we do o/n 01 q7??????? is dere any oder method apart frm da 1 in solved pprx???
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Dude
u have to do
1/6 ----Beacuse out 6 paper clip in Box A They r asking probablity for picking Red
7/10 ----Beacuse out of 10 paper clip in box B Then r asking probablity of getting Reds will b in 7/10 ways
Therefore
Total Probablity of this Sub-part of the question is [(1/6 X 7/10)]
=(7/60)