Qualification > Math
Urgent: Maths CIE Stats
Deadly_king:
--- Quote from: mousa on October 24, 2010, 05:56:30 pm ---Wowww nice, muritaniaaaa loool, thats tooo far from here :P I am Jordanian living in the UAE, so you are Muratious ya3ny??
--- End quote ---
Hehe........Not Mauritania but Mauritius, tiny island in the Indian Ocean :D
$!$RatJumper$!$:
--- Quote from: mousa on October 22, 2010, 06:02:57 am ---I would like to add more, the probability of success in " choosing a red sweet " is NOT constant, it keeps on changing everytime you make a selection since the sweets are not replaced. Furthermore, the trials aren't independent. Both of those conditions are essential for binomial distribution.
hope that helped, by the way, your doing your exam in this nov session??
--- End quote ---
When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.
Deadly_king:
--- Quote from: $!$RatJumper$!$ on October 26, 2010, 05:52:19 pm ---When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.
--- End quote ---
The trials are NOT independent ----> each trial depends on the previous trial since the previous trial decreased the total number of sweets in the bag, so the probability will depend on how many sweets have been removed and have they been replaced or not. ;)
cs:
Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2
why 6! X 7C3?
Deadly_king:
--- Quote from: cs on October 29, 2010, 01:25:34 pm ---Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2
why 6! X 7C3?
--- End quote ---
I can't really explain the workings from the marking scheme but I did it another way which was approved by my teacher.
First we find out the number of arrangements in which all the 9 persons can be arranged. This will be (9! +(7! x 3!))
9! => There are nine persons which can be arranged in any way.
(7! x 3!) => There are 6 men and 3 women. We'll be taking all the women to be standing beside each other. Hence they are taken as one possiblity only which makes 7!. The 3 women can then be arranged in any way 3!.
Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
8! = > One woman is taken as one possiblity and the other two as another possiblity. In all there will be 8 possiblities, after adding the six men.
Number of arrangements : (9! +(7! x 3!)) - 3(8! x 2!) = 151200
Hope you understand :)
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