STATISTICS

[preity]

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.....

[astarmathsandphysics]

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

[astarmathsandphysics]

I meant 5/25 for b since 5 out the 25 that own a car also own a bike

[preity]

thanks ya....

[ocean]

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.

[louis]

(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 )
          -------------------------------------------------------------------
          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)
                        --------------------------------------------------------------------------------------------------
                                                n (person from both countries A and B who drink  tea)

       = (3X 0.7 )  +  (12 X 0.35)
          -----------------------------   
                        15

      =  2.1  +  4.2   
          ------------
                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
     -------------------
           0.42
=     0.667


     
       
         

[candy]

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!

[preity]

Candy...mayb u should try this website...http://www.s-cool.co.uk/alevel/maths.html

[me@me@me]

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

[candy]

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?

[louis]

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

[louis]

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.

[preity]

thanks a lot louis...

[candy]

(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 

[mac]

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???