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STATISTICS

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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???  ???

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